Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://www.codechef.com/ENDE2020/problems/ENCDEC2:
You are the principal of the Cake school in chefland and today is your birthday. You want to treat each of the children with a small cupcake which is made by you. But there is a problem, You don't know how many students are present today.
The students have gathered of the morning assembly in $R$ rows and $C$ columns. Now you have to calculate how many cakes you have to make such that each child gets a cupcake. 

-----Input:-----
- First-line will contain $T$, the number of test cases. Then the test cases follow. 
- Each test case contains a single line of input, two integers $R$ and $C$. 

-----Output:-----
For each test case, output number of cupcakes you have to make.

-----Constraints-----
- $1 \leq T \leq 1000$
- $2 \leq R,C \leq 10^6$

-----Sample Input:-----
1
5 10

-----Sample Output:-----
50

I tried it in Python, but could not do it. Can you solve it?
targets: # cook your dish here
t=int(input())
for i in range(t):
    l=list(map(int,input().split()))
    r=l[0]
    c=l[1]
    print(r*c)
inputs: Solve in Python:
Simon has a prime number x and an array of non-negative integers a_1, a_2, ..., a_{n}.

Simon loves fractions very much. Today he wrote out number $\frac{1}{x^{a} 1} + \frac{1}{x^{a_{2}}} + \ldots + \frac{1}{x^{a_{n}}}$ on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: $\frac{s}{t}$, where number t equals x^{a}_1 + a_2 + ... + a_{n}. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (10^9 + 7).


-----Input-----

The first line contains two positive integers n and x (1 ≤ n ≤ 10^5, 2 ≤ x ≤ 10^9) — the size of the array and the prime number.

The second line contains n space-separated integers a_1, a_2, ..., a_{n} (0 ≤ a_1 ≤ a_2 ≤ ... ≤ a_{n} ≤ 10^9). 


-----Output-----

Print a single number — the answer to the problem modulo 1000000007 (10^9 + 7).


-----Examples-----
Input
2 2
2 2

Output
8

Input
3 3
1 2 3

Output
27

Input
2 2
29 29

Output
73741817

Input
4 5
0 0 0 0

Output
1



-----Note-----

In the first sample $\frac{1}{4} + \frac{1}{4} = \frac{4 + 4}{16} = \frac{8}{16}$. Thus, the answer to the problem is 8.

In the second sample, $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{243 + 81 + 27}{729} = \frac{351}{729}$. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample $\frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = \frac{4}{1}$. Thus, the answer to the problem is 1.
targets: n, x = map(int, input().split())
t = list(map(int, input().split()))
b = max(t)
k = t.count(b)
r = sum(t)
s = r - b
while k % x == 0:
    b -= 1
    s += 1
    k = k // x + t.count(b)
print(pow(x, min(r, s), 1000000007))
inputs: Solve in Python:
During the quarantine, Sicromoft has more free time to create the new functions in "Celex-2021". The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:

 [Image]  The cell with coordinates $(x, y)$ is at the intersection of $x$-th row and $y$-th column. Upper left cell $(1,1)$ contains an integer $1$.

The developers of the SUM function don't sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell $(x,y)$ in one step you can move to the cell $(x+1, y)$ or $(x, y+1)$. 

After another Dinwows update, Levian started to study "Celex-2021" (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell $(x_1, y_1)$ to another given cell $(x_2, y_2$), if you can only move one cell down or right.

Formally, consider all the paths from the cell $(x_1, y_1)$ to cell $(x_2, y_2)$ such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 57179$) — the number of test cases.

Each of the following $t$ lines contains four natural numbers $x_1$, $y_1$, $x_2$, $y_2$ ($1 \le x_1 \le x_2 \le 10^9$, $1 \le y_1 \le y_2 \le 10^9$) — coordinates of the start and the end cells. 


-----Output-----

For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.


-----Example-----
Input
4
1 1 2 2
1 2 2 4
179 1 179 100000
5 7 5 7

Output
2
3
1
1



-----Note-----

In the first test case there are two possible sums: $1+2+5=8$ and $1+3+5=9$. [Image]
targets: import sys
input = sys.stdin.readline

def value(x,y):
    ANS=(x+y-2)*(x+y-1)//2
    return ANS+x

t=int(input())
for tests in range(t):
    x1,y1,x2,y2=list(map(int,input().split()))

    print(1+(x2-x1)*(y2-y1))
inputs: Solve in Python:
Recently, Olya received a magical square with the size of $2^n\times 2^n$.

It seems to her sister that one square is boring. Therefore, she asked Olya to perform exactly $k$ splitting operations.

A Splitting operation is an operation during which Olya takes a square with side $a$ and cuts it into 4 equal squares with side $\dfrac{a}{2}$. If the side of the square is equal to $1$, then it is impossible to apply a splitting operation to it (see examples for better understanding).

Olya is happy to fulfill her sister's request, but she also wants the condition of Olya's happiness to be satisfied after all operations.

The condition of Olya's happiness will be satisfied if the following statement is fulfilled:

Let the length of the side of the lower left square be equal to $a$, then the length of the side of the right upper square should also be equal to $a$. There should also be a path between them that consists only of squares with the side of length $a$. All consecutive squares on a path should have a common side.

Obviously, as long as we have one square, these conditions are met. So Olya is ready to fulfill her sister's request only under the condition that she is satisfied too. Tell her: is it possible to perform exactly $k$ splitting operations in a certain order so that the condition of Olya's happiness is satisfied? If it is possible, tell also the size of the side of squares of which the path from the lower left square to the upper right one will consist.


-----Input-----

The first line contains one integer $t$ ($1 \le t \le 10^3$) — the number of tests.

Each of the following $t$ lines contains two integers $n_i$ and $k_i$ ($1 \le n_i \le 10^9, 1 \le k_i \le 10^{18}$) — the description of the $i$-th test, which means that initially Olya's square has size of $2^{n_i}\times 2^{n_i}$ and Olya's sister asks her to do exactly $k_i$ splitting operations.


-----Output-----

Print $t$ lines, where in the $i$-th line you should output "YES" if it is possible to perform $k_i$ splitting operations in the...
targets: import sys
import math
input = sys.stdin.readline

testcase=int(input())
T=[list(map(int,input().split())) for i in range(testcase)]

def bi(n,k):
    MIN=0
    MAX=n

    while MAX>MIN+1:
        bn=(MIN+MAX)//2
        if math.log2(k+2+bn)<bn+1:
            MAX=bn
        elif math.log2(k+2+bn)==bn+1:
            return bn
        else:
            MIN=bn

    if MAX+1<=math.log2(k+2+MAX):
        return MAX

    return MIN

for n,k in T:
    if n==1:
        if k==1:
            print("YES",0)
        else:
            print("NO")
        continue

    if n==2:
        if 1<=k<=2:
            print("YES",1)
        elif k==3:
            print("NO")
        elif 4<=k<=5:
            print("YES",0)
        else:
            print("NO")
        continue
        
    if n<=30 and k>(pow(4,n)-1)//3:
        print("NO")
        continue

    ANS=bi(n,k)

    print("YES",n-ANS)
inputs: Solve in Python:
Recently you have received two positive integer numbers $x$ and $y$. You forgot them, but you remembered a shuffled list containing all divisors of $x$ (including $1$ and $x$) and all divisors of $y$ (including $1$ and $y$). If $d$ is a divisor of both numbers $x$ and $y$ at the same time, there are two occurrences of $d$ in the list.

For example, if $x=4$ and $y=6$ then the given list can be any permutation of the list $[1, 2, 4, 1, 2, 3, 6]$. Some of the possible lists are: $[1, 1, 2, 4, 6, 3, 2]$, $[4, 6, 1, 1, 2, 3, 2]$ or $[1, 6, 3, 2, 4, 1, 2]$.

Your problem is to restore suitable positive integer numbers $x$ and $y$ that would yield the same list of divisors (possibly in different order).

It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers $x$ and $y$.


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 128$) — the number of divisors of $x$ and $y$.

The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^4$), where $d_i$ is either divisor of $x$ or divisor of $y$. If a number is divisor of both numbers $x$ and $y$ then there are two copies of this number in the list.


-----Output-----

Print two positive integer numbers $x$ and $y$ — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.


-----Example-----
Input
10
10 2 8 1 2 4 1 20 4 5

Output
20 8
targets: n = int(input())
a = list(map(int, input().split()))
x = max(a)
b = []
for i in range(1, x + 1) :
    if x % i == 0 :
        b.append(i)

for i in range(len(b)) :
    a.remove(b[i])

print(x, max(a))
inputs: I found an interesting problem on https://www.codechef.com/problems/ADTRI:
Rupsa really loves triangles. One day she came across an equilateral triangle having length of each side as an integer N. She started wondering if it was possible to transform the triangle keeping two sides fixed and alter the third side such that it still remains a triangle, but the altered side will have its length as an even integer, and the line drawn from the opposite vertex to the mid-point of the altered side is of integral length.
Since Rupsa is in a hurry to record a song for Chef as he really loves her songs, you must help her solve the problem as fast as possible.

-----Input-----
The first line of input contains an integer T denoting the number of test cases.
Each test-case contains a single integer N.

-----Output-----

For each test case, output "YES" if the triangle transformation is possible, otherwise "NO" (quotes for clarity only, do not output).

-----Constraints-----
- 1 ≤ T ≤ 106
- 1 ≤ N ≤ 5 x 106

-----Sub tasks-----
- Subtask #1: 1 ≤ T ≤ 100, 1 ≤ N ≤ 104 (10 points)
- Subtask #2: 1 ≤ T ≤ 104, 1 ≤ N ≤ 106 (30 points)
- Subtask #3: Original Constraints (60 points)

-----Example-----
Input:2
5
3

Output:YES
NO

-----Explanation-----
- In test case 1, make the length of any one side 6, and it will suffice.

I tried it in Python, but could not do it. Can you solve it?
targets: t=int(input())
import math

for i in range(t):
 b=[]
 n=int(input())
 for p in range(n+1,2*n):
  if p%2==0:
   
   a= math.sqrt(n**2 - (p/2)**2)
   if a-int(a)==0:
    b.append(i)
 if len(b)==0:
  print("NO")
 elif len(b)!=0:
  print("YES")
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1392/F:
Omkar is standing at the foot of Celeste mountain. The summit is $n$ meters away from him, and he can see all of the mountains up to the summit, so for all $1 \leq j \leq n$ he knows that the height of the mountain at the point $j$ meters away from himself is $h_j$ meters. It turns out that for all $j$ satisfying $1 \leq j \leq n - 1$, $h_j < h_{j + 1}$ (meaning that heights are strictly increasing).

Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if $h_j + 2 \leq h_{j + 1}$, then one square meter of dirt will slide from position $j + 1$ to position $j$, so that $h_{j + 1}$ is decreased by $1$ and $h_j$ is increased by $1$. These changes occur simultaneously, so for example, if $h_j + 2 \leq h_{j + 1}$ and $h_{j + 1} + 2 \leq h_{j + 2}$ for some $j$, then $h_j$ will be increased by $1$, $h_{j + 2}$ will be decreased by $1$, and $h_{j + 1}$ will be both increased and decreased by $1$, meaning that in effect $h_{j + 1}$ is unchanged during that minute.

The landslide ends when there is no $j$ such that $h_j + 2 \leq h_{j + 1}$. Help Omkar figure out what the values of $h_1, \dots, h_n$ will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.

Note that because of the large amount of input, it is recommended that your code uses fast IO.


-----Input-----

The first line contains a single integer $n$ ($1 \leq n \leq 10^6$). 

The second line contains $n$ integers $h_1, h_2, \dots, h_n$ satisfying $0 \leq h_1 < h_2 < \dots < h_n \leq 10^{12}$ — the heights.


-----Output-----

Output $n$ integers, where the $j$-th integer is the value of $h_j$ after the landslide has stopped.


-----Example-----
Input
4
2 6 7 8

Output
5 5 6 7



-----Note-----

Initially, the mountain has heights $2, 6, 7, 8$.

In the first minute, we have $2 + 2 \leq 6$, so $2$ increases to $3$ and $6$ decreases to $5$, leaving $3, 5, 7, 8$.

In the second minute, we have $3 + 2 \leq 5$ and $5 + 2 \leq 7$, so...

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input());tot = sum(map(int, input().split()));extra = (n * (n - 1))//2;smol = (tot - extra) // n;out = [smol + i for i in range(n)]
for i in range(tot - sum(out)):out[i] += 1
print(' '.join(map(str,out)))
inputs: I found an interesting problem on https://www.codewars.com/kata/5ae8e14c1839f14b2c00007a:
# Idea

In the world of graphs exists a structure called "spanning tree". It is unique because it's created not on its own, but based on other graphs. To make a spanning tree out of a given graph you should remove all the edges which create cycles, for example:
```
This can become      this      or      this        or       this

 A                    A                  A                   A
 |\                   |                   \                  |\
 | \       ==>        |                    \                 | \
 |__\                 |__                 __\                |  \
B   C                 B  C               B  C                B  C
```
Each *edge* (line between 2 *vertices*, i.e. points) has a weight, based on which you can build minimum and maximum spanning trees (sum of weights of vertices in the resulting tree is minimal/maximal possible).  
[Wikipedia article](https://en.wikipedia.org/wiki/Spanning_tree) on spanning trees, in case you need it.

___

# Task

You will receive an array like this: `[["AB", 2], ["BC", 4], ["AC", 1]]` which includes all edges of an arbitrary graph and a string `"min"`/`"max"`. Based on them you should get and return a new array which includes only those edges which form a minimum/maximum spanning trees.
```python
edges = [("AB", 2), ("BC", 4), ("AC", 1)]

make_spanning_tree(edges, "min")    ==>    [("AB", 2), ("AC", 1)]
make_spanning_tree(edges, "max")    ==>    [("AB", 2), ("BC", 4)]
```

___

# Notes

* All vertices will be connected with each other
* You may receive cycles, for example - `["AA", n]`
* The subject of the test are these 3 values: number of vertices included, total weight, number of edges, but **you should not return them**, there's a special function which will analyze your output instead

I tried it in Python, but could not do it. Can you solve it?
targets: def make_spanning_tree(edges, t):
    edges.sort(reverse=True)
    last_node = ord(edges[0][0][0]) - 65
    edges.sort()
    edges.sort(key=lambda edges: edges[1])

    if(t == 'max'):
        edges.reverse()
    
    nonlocal parent
    parent = [int(x) for x in range(last_node+10000)]
    result = []

    for edge in edges:
        startNode = ord(edge[0][0])-65
        endnode = ord(edge[0][1])-65

        if(Find(startNode) != Find(endnode)):
            Union(startNode, endnode)
            result.append(edge)

    return result

def Find(x):
    if(x == parent[x]):
        return x
    else:
        p = Find(parent[x])
        parent[x] = p
        return p

def Union(x, y):
    x = Find(x)
    y = Find(y)

    if(x != y):
        parent[y] = x
inputs: Solve in Python:
Consider a football tournament where n teams participate. Each team has two football kits: for home games, and for away games. The kit for home games of the i-th team has color x_{i} and the kit for away games of this team has color y_{i} (x_{i} ≠ y_{i}).

In the tournament, each team plays exactly one home game and exactly one away game with each other team (n(n - 1) games in total). The team, that plays the home game, traditionally plays in its home kit. The team that plays an away game plays in its away kit. However, if two teams has the kits of the same color, they cannot be distinguished. In this case the away team plays in its home kit.

Calculate how many games in the described tournament each team plays in its home kit and how many games it plays in its away kit.


-----Input-----

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of teams. Next n lines contain the description of the teams. The i-th line contains two space-separated numbers x_{i}, y_{i} (1 ≤ x_{i}, y_{i} ≤ 10^5; x_{i} ≠ y_{i}) — the color numbers for the home and away kits of the i-th team.


-----Output-----

For each team, print on a single line two space-separated integers — the number of games this team is going to play in home and away kits, correspondingly. Print the answers for the teams in the order they appeared in the input.


-----Examples-----
Input
2
1 2
2 1

Output
2 0
2 0

Input
3
1 2
2 1
1 3

Output
3 1
4 0
2 2
targets: n = int( input() )
x, y = [0] * n, [0] * n
for i in range( n ):
    x[i], y[i] = map( int, input().split() )
ans = [0] * 100001
for elm in x:
    ans[elm] += 1
for elm in y:
    print( n - 1 + ans[elm], n - 1 - ans[elm] )
inputs: Solve in Python:
In this kata, you have to define a function named **func** that will take a list as input.

You must try and guess the pattern how we get the output number and return list - **[output number,binary representation,octal representation,hexadecimal representation]**, but **you must convert that specific number without built-in : bin,oct and hex functions.**

Examples : 

```python
func([12,13,6,3,6,45,123]) returns - [29,'11101','35','1d']

func([1,9,23,43,65,31,63,99]) returns - [41,'101001','51','29']

func([2,4,6,8,10,12,14,16,18,19]) returns - [10,'1010','12','a']
```
targets: def binary(x):
    result = []
    while True:
        remainder = x % 2
        result.append(str(remainder))
        x = x // 2

        if x == 0:
            break

    result.reverse()
    return "".join(result)

def octal(x):
    result = []
    while True:
        remainder = x % 8
        result.append(str(remainder))
        x = x // 8

        if x == 0:
            break

    result.reverse()
    return "".join(result)

def hexadecimal(x):
    result = []
    while True:
        remainder = x % 16

        if remainder == 10:
            result.append('a')
        elif remainder == 11:
            result.append('b')
        elif remainder == 12:
            result.append('c')
        elif remainder == 13:
            result.append('d')
        elif remainder == 14:
            result.append('e')
        elif remainder == 15:
            result.append('f')
        else:
            result.append(str(remainder))

        x = x // 16
        if x == 0:
            break

    result.reverse()
    return "".join(result)

import math

def func(l):
    average = math.floor(sum(l) / len(l))
    result = [average, binary(average), octal(average), hexadecimal(average)]
    return result
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/424/A:
Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.

For another exercise, Pasha needs exactly $\frac{n}{2}$ hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?


-----Input-----

The first line contains integer n (2 ≤ n ≤ 200; n is even). The next line contains n characters without spaces. These characters describe the hamsters' position: the i-th character equals 'X', if the i-th hamster in the row is standing, and 'x', if he is sitting.


-----Output-----

In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.


-----Examples-----
Input
4
xxXx

Output
1
XxXx

Input
2
XX

Output
1
xX

Input
6
xXXxXx

Output
0
xXXxXx

I tried it in Python, but could not do it. Can you solve it?
targets: n, t = int(input()), input()
a, b = 'x', 'X'
k = t.count(a) - n // 2
if k:
    t = list(t)
    if k < 0: a, b, k = b, a, -k
    print(k)
    for i, c in enumerate(t):
        if c == a:
            t[i] = b
            k -= 1
            if k == 0: break
    print(''.join(t))
else: print('0\n' + t)
inputs: I found an interesting problem on https://www.codewars.com/kata/5c765a4f29e50e391e1414d4:
[Haikus](https://en.wikipedia.org/wiki/Haiku_in_English) are short poems in a three-line format, with 17 syllables arranged in a 5–7–5 pattern. Your task is to check if the supplied text is a haiku or not.


### About syllables

[Syllables](https://en.wikipedia.org/wiki/Syllable) are the phonological building blocks of words. *In this kata*, a syllable is a part of a word including a vowel ("a-e-i-o-u-y") or a group of vowels (e.g. "ou", "ee", "ay"). A few examples: "tea", "can", "to·day", "week·end", "el·e·phant".

**However**, silent "E"s **do not** create syllables. *In this kata*, an "E" is considered silent if it's alone at the end of the word, preceded by one (or more) consonant(s) and there is at least one other syllable in the word. Examples: "age", "ar·range", "con·crete"; but not in "she", "blue", "de·gree".

Some more examples:
* one syllable words: "cat", "cool", "sprout", "like", "eye", "squeeze"
* two syllables words: "ac·count", "hon·est", "beau·ty", "a·live", "be·cause", "re·store"


## Examples
```
An old silent pond...
A frog jumps into the pond,
splash! Silence again.
```

...should return `True`, as this is a valid 5–7–5 haiku:
```
An old si·lent pond...            # 5 syllables
A frog jumps in·to the pond,      # 7
splash! Si·lence a·gain.          # 5
```

Another example:

```
Autumn moonlight -
a worm digs silently
into the chestnut.
```

...should return `False`, because the number of syllables per line is not correct:
```
Au·tumn moon·light -          # 4 syllables
a worm digs si·lent·ly        # 6
in·to the chest·nut.          # 5
```

---

## My other katas

If you enjoyed this kata then please try [my other katas](https://www.codewars.com/collections/katas-created-by-anter69)! :-)

I tried it in Python, but could not do it. Can you solve it?
targets: def is_haiku(s,v='aeiouy'):
    C=[]
    for x in[a.strip('.;:!?,').split()for a in s.lower().split('\n')]:
        t=''
        for y in x:
            if y.endswith('e')and any(c in y[:-1]for c in v):y=y[:-1]
            for c in y:t+=c if c in v else' 'if t and t[-1]!=' 'else''
            t+=' '
        C+=[len(t.split())]
    return C==[5,7,5]
inputs: Solve in Python:
Takahashi has K 500-yen coins. (Yen is the currency of Japan.)
If these coins add up to X yen or more, print Yes; otherwise, print No.

-----Constraints-----
 - 1 \leq K \leq 100
 - 1 \leq X \leq 10^5

-----Input-----
Input is given from Standard Input in the following format:
K X

-----Output-----
If the coins add up to X yen or more, print Yes; otherwise, print No.

-----Sample Input-----
2 900

-----Sample Output-----
Yes

Two 500-yen coins add up to 1000 yen, which is not less than X = 900 yen.
targets: a,b=input().split();print('YNeos'[eval(a+'*500<'+b)::2])
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1108/B:
Recently you have received two positive integer numbers $x$ and $y$. You forgot them, but you remembered a shuffled list containing all divisors of $x$ (including $1$ and $x$) and all divisors of $y$ (including $1$ and $y$). If $d$ is a divisor of both numbers $x$ and $y$ at the same time, there are two occurrences of $d$ in the list.

For example, if $x=4$ and $y=6$ then the given list can be any permutation of the list $[1, 2, 4, 1, 2, 3, 6]$. Some of the possible lists are: $[1, 1, 2, 4, 6, 3, 2]$, $[4, 6, 1, 1, 2, 3, 2]$ or $[1, 6, 3, 2, 4, 1, 2]$.

Your problem is to restore suitable positive integer numbers $x$ and $y$ that would yield the same list of divisors (possibly in different order).

It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers $x$ and $y$.


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 128$) — the number of divisors of $x$ and $y$.

The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^4$), where $d_i$ is either divisor of $x$ or divisor of $y$. If a number is divisor of both numbers $x$ and $y$ then there are two copies of this number in the list.


-----Output-----

Print two positive integer numbers $x$ and $y$ — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.


-----Example-----
Input
10
10 2 8 1 2 4 1 20 4 5

Output
20 8

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())
a = list(map(int, input().split()))
x = max(a)
b = []
for i in range(1, x + 1) :
    if x % i == 0 :
        b.append(i)

for i in range(len(b)) :
    a.remove(b[i])

print(x, max(a))
inputs: Solve in Python:
The only difference between easy and hard versions is the number of elements in the array.

You are given an array $a$ consisting of $n$ integers. In one move you can choose any $a_i$ and divide it by $2$ rounding down (in other words, in one move you can set $a_i := \lfloor\frac{a_i}{2}\rfloor$).

You can perform such an operation any (possibly, zero) number of times with any $a_i$.

Your task is to calculate the minimum possible number of operations required to obtain at least $k$ equal numbers in the array.

Don't forget that it is possible to have $a_i = 0$ after some operations, thus the answer always exists.


-----Input-----

The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2 \cdot 10^5$) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$), where $a_i$ is the $i$-th element of $a$.


-----Output-----

Print one integer — the minimum possible number of operations required to obtain at least $k$ equal numbers in the array.


-----Examples-----
Input
5 3
1 2 2 4 5

Output
1

Input
5 3
1 2 3 4 5

Output
2

Input
5 3
1 2 3 3 3

Output
0
targets: from collections import *
n,k = [int(i) for i in input().split()]
A = [int(i) for i in input().split()]
A.sort()
d = defaultdict(lambda: [0,0])
for i in A:
	num = i
	stp = 0
	while num>0:
		if d[num][0]<k:
			d[num][0]+=1
			d[num][1]+=stp
		num = num//2
		stp +=1
	if d[num][0]<k:
			d[num][0]+=1
			d[num][1]+=stp

mn = 10**20
for i in d:
	if(d[i][0]>=k):
		mn = min(d[i][1],mn)

print(mn)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/445/B:
DZY loves chemistry, and he enjoys mixing chemicals.

DZY has n chemicals, and m pairs of them will react. He wants to pour these chemicals into a test tube, and he needs to pour them in one by one, in any order. 

Let's consider the danger of a test tube. Danger of an empty test tube is 1. And every time when DZY pours a chemical, if there are already one or more chemicals in the test tube that can react with it, the danger of the test tube will be multiplied by 2. Otherwise the danger remains as it is.

Find the maximum possible danger after pouring all the chemicals one by one in optimal order.


-----Input-----

The first line contains two space-separated integers n and m $(1 \leq n \leq 50 ; 0 \leq m \leq \frac{n(n - 1)}{2})$.

Each of the next m lines contains two space-separated integers x_{i} and y_{i} (1 ≤ x_{i} < y_{i} ≤ n). These integers mean that the chemical x_{i} will react with the chemical y_{i}. Each pair of chemicals will appear at most once in the input.

Consider all the chemicals numbered from 1 to n in some order.


-----Output-----

Print a single integer — the maximum possible danger.


-----Examples-----
Input
1 0

Output
1

Input
2 1
1 2

Output
2

Input
3 2
1 2
2 3

Output
4



-----Note-----

In the first sample, there's only one way to pour, and the danger won't increase.

In the second sample, no matter we pour the 1st chemical first, or pour the 2nd chemical first, the answer is always 2.

In the third sample, there are four ways to achieve the maximum possible danger: 2-1-3, 2-3-1, 1-2-3 and 3-2-1 (that is the numbers of the chemicals in order of pouring).

I tried it in Python, but could not do it. Can you solve it?
targets: n, m = map(int, input().split())
p = list(range(n + 1))
s = [{i} for i in p]
for i in range(m):
    x, y = map(int, input().split())
    x, y = p[x], p[y]
    if x != y:
        if len(s[x]) < len(s[y]):
            for k in s[x]:
                p[k] = y
                s[y].add(k)
            s[x].clear()
        else:
            for k in s[y]:
                p[k] = x
                s[x].add(k)
            s[y].clear()
q = 0
for p in s:
    if p: q += len(p) - 1
print(1 << q)
inputs: Solve in Python:
Alice and Bob have participated to a Rock Off with their bands. A jury of true metalheads rates the two challenges, awarding points to the bands on a scale from 1 to 50 for three categories: Song Heaviness, Originality, and Members' outfits.

For each one of these 3 categories they are going to be awarded one point, should they get a better judgement from the jury. No point is awarded in case of an equal vote.

You are going to receive two arrays, containing first the score of Alice's band and then those of Bob's. Your task is to find their total score by comparing them in a single line. 

Example:

Alice's band plays a Nirvana inspired grunge and has been rated 
``20`` for Heaviness, 
``32`` for Originality and only 
``18`` for Outfits.
Bob listens to Slayer and has gotten a good 
``48`` for Heaviness, 
``25`` for Originality and a rather honest 
``40`` for Outfits.

The total score should be followed by a colon ```:``` and by one of the following quotes:
if Alice's band wins: ```Alice made "Kurt" proud!```
if Bob's band wins: ```Bob made "Jeff" proud!```
if they end up with a draw: ```that looks like a "draw"! Rock on!```

The solution to the example above should therefore appear like 
``'1, 2: Bob made "Jeff" proud!'``.
targets: def solve(a, b):
    count_a = 0
    count_b = 0
    for i in range(len(a)):
        if a[i] > b[i]:
            count_a +=1
        elif a[i] < b[i]:
            count_b +=1
    return f'{count_a}, {count_b}: that looks like a "draw"! Rock on!' if count_a == count_b else f'{count_a}, {count_b}: Alice made "Kurt" proud!' if count_a > count_b else f'{count_a}, {count_b}: Bob made "Jeff" proud!'
inputs: Solve in Python:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:  if the last digit of the number is non-zero, she decreases the number by one;  if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit). 

You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.

It is guaranteed that the result will be positive integer number.


-----Input-----

The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.


-----Output-----

Print one integer number — the result of the decreasing $n$ by one $k$ times.

It is guaranteed that the result will be positive integer number. 


-----Examples-----
Input
512 4

Output
50

Input
1000000000 9

Output
1



-----Note-----

The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$.
targets: n,k = list(map(int,input().split()))
for i in range(k):
    if n%10 == 0: n//=10
    else: n -=1
print(n)
inputs: Solve in Python:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings a and b of equal length are called equivalent in one of the two cases:   They are equal.  If we split string a into two halves of the same size a_1 and a_2, and string b into two halves of the same size b_1 and b_2, then one of the following is correct:   a_1 is equivalent to b_1, and a_2 is equivalent to b_2  a_1 is equivalent to b_2, and a_2 is equivalent to b_1  

As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.

Gerald has already completed this home task. Now it's your turn!


-----Input-----

The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200 000 and consists of lowercase English letters. The strings have the same length.


-----Output-----

Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.


-----Examples-----
Input
aaba
abaa

Output
YES

Input
aabb
abab

Output
NO



-----Note-----

In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".

In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
targets: def F(s):
    if len(s)%2==1:return s
    s1 = F(s[:len(s)//2])
    s2 = F(s[len(s)//2:])
    if s1 < s2:return s1 + s2
    return s2 + s1
if F(input()) == F(input()):
    print("YES")
else:
    print("NO")
inputs: Solve in Python:
Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:  for the array $[0, 0, 1, 0, 2]$ MEX equals to $3$ because numbers $0, 1$ and $2$ are presented in the array and $3$ is the minimum non-negative integer not presented in the array;  for the array $[1, 2, 3, 4]$ MEX equals to $0$ because $0$ is the minimum non-negative integer not presented in the array;  for the array $[0, 1, 4, 3]$ MEX equals to $2$ because $2$ is the minimum non-negative integer not presented in the array. 

You are given an empty array $a=[]$ (in other words, a zero-length array). You are also given a positive integer $x$.

You are also given $q$ queries. The $j$-th query consists of one integer $y_j$ and means that you have to append one element $y_j$ to the array. The array length increases by $1$ after a query.

In one move, you can choose any index $i$ and set $a_i := a_i + x$ or $a_i := a_i - x$ (i.e. increase or decrease any element of the array by $x$). The only restriction is that $a_i$ cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of $q$ queries (i.e. the $j$-th answer corresponds to the array of length $j$).

Operations are discarded before each query. I.e. the array $a$ after the $j$-th query equals to $[y_1, y_2, \dots, y_j]$.


-----Input-----

The first line of the input contains two integers $q, x$ ($1 \le q, x \le 4 \cdot 10^5$) — the number of queries and the value of $x$.

The next $q$ lines describe queries. The $j$-th query consists of one integer $y_j$ ($0 \le y_j \le 10^9$) and means that you have to append one element $y_j$ to the array.


-----Output-----

Print the answer to the initial problem after each query — for the query $j$ print the maximum value of MEX after first $j$...
targets: import sys
input = sys.stdin.readline

q,x=list(map(int,input().split()))

seg_el=1<<((x+1).bit_length())
SEG=[0]*(2*seg_el)

def add1(n,seg_el):
    i=n+seg_el
    SEG[i]+=1
    i>>=1
    
    while i!=0:
        SEG[i]=min(SEG[i*2],SEG[i*2+1])
        i>>=1
        
def getvalues(l,r):
    L=l+seg_el
    R=r+seg_el
    ANS=1<<30

    while L<R:
        if L & 1:
            ANS=min(ANS , SEG[L])
            L+=1

        if R & 1:
            R-=1
            ANS=min(ANS , SEG[R])
        L>>=1
        R>>=1

    return ANS

for qu in range(q):
    t=int(input())
    add1(t%x,seg_el)

    #print(SEG)

    MIN=getvalues(0,x)

    OK=0
    NG=x

    while NG-OK>1:
        mid=(OK+NG)//2

        if getvalues(0,mid)>MIN:
            OK=mid
        else:
            NG=mid

    #print(MIN,OK)
    print(MIN*x+OK)
inputs: Solve in Python:
You are given a tree with $N$ vertices (numbered $1$ through $N$) and a bag with $N$ markers. There is an integer written on each marker; each of these integers is $0$, $1$ or $2$. You must assign exactly one marker to each vertex.
Let's define the unattractiveness of the resulting tree as the maximum absolute difference of integers written on the markers in any two vertices which are connected by an edge.
Find the minimum possible unattractiveness of the resulting tree.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains a single integer $N$.
- The second line contains $N$ space-separated integers denoting the numbers on markers in the bag.
- Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting an edge between vertices $u$ and $v$.

-----Output-----
For each test case, print a single line containing one integer — the minimum unattractiveness.

-----Constraints-----
- $1 \le T \le 30$
- $1 \le N \le 100$
- $1 \le u, v \le N$
- the graph described on the input is a tree

-----Example Input-----
3
3
0 1 1
1 2
1 3
3
0 1 2
1 2
1 3
4
2 2 2 2
1 2
1 3
3 4

-----Example Output-----
1
1
0

-----Explanation-----
Example case 1:
targets: # cook your dish here
import numpy as np
tests = int(input())
for _ in range(tests):
   n = int(input())
   weights = [int(j) for j in input().split()]
   edges = [[0] for _ in range(n-1)]
   for i in range(n-1):
      edges[i] = [int(j)-1 for j in input().split()]
   vertex_set = [[] for _ in range(n)]
   for i in range(n-1):
      vertex_set[edges[i][0]].append(edges[i][1])
      vertex_set[edges[i][1]].append(edges[i][0])
   counts = [0 for _ in range(3)]
   for i in range(n):
      counts[weights[i]] += 1
   if counts[1] == 0:
      print(2 * (counts[0] != 0 and counts[2] != 0))
   elif counts[1] == n:
      print(0)
   else:
      visited = [0]
      for i in range(n):
         vertex = visited[i]
         for v in vertex_set[vertex]:
            if v not in visited:
               visited.append(v)
      vertex_nums = [[0] for _ in range(n)]
      for i in range(n-1,-1,-1):
         vertex = visited[i]
         for v in vertex_set[vertex]:
            if v in visited[i:]:
               vertex_nums[vertex].append(sum(vertex_nums[v])+1)
      for i in range(n):
         vertex_nums[i].append(n-1-sum(vertex_nums[i]))
      sums = np.zeros(n,dtype=bool)
      sums[0] = True
      for i in range(n):
         new_sums = np.zeros(n,dtype=bool)
         new_sums[0] = True
         for num in vertex_nums[i]:
            new_sums[num:n] = np.logical_or(new_sums[num:n],new_sums[:n-num])
         sums = np.logical_or(sums,new_sums)
      solved = False
      for i in range(n):
         if sums[i] and counts[0] <= i and counts[2] <= n - 1 - i:
            solved = True
            break
      if solved or counts[1] > 1:
         print(1)
      else:
         print(2)
inputs: I found an interesting problem on https://www.codewars.com/kata/5416f1834c24604c46000696:
In computer science, cycle detection is the algorithmic problem of finding a cycle in a sequence of iterated function values.

For any function ƒ, and any initial value x0 in S, the sequence of iterated function values

    x0,x1=f(x0), x2=f(x1), ...,xi=f(x{i-1}),...

may eventually use the same value twice under some assumptions: S finite, f periodic ... etc. So there will be some `i ≠ j` such that `xi = xj`. Once this happens, the sequence must continue by repeating the cycle of values from `xi to xj−1`. Cycle detection is the problem of finding `i` and `j`, given `ƒ` and `x0`. Let `μ` be the smallest index such that the value associated will reappears and `λ` the smallest value such that `xμ = xλ+μ, λ` is the loop length.

Example:

Consider the sequence: 
```
2, 0, 6, 3, 1, 6, 3, 1, 6, 3, 1, ....
```

The cycle in this value sequence is 6, 3, 1.
μ is 2 (first 6)
λ is 3 (length of the sequence or difference between position of consecutive 6).

The goal of this kata is to build a function that will return `[μ,λ]` when given a short sequence. Simple loops will be sufficient. The sequence will be given in the form of an array. All array will be valid sequence associated with deterministic function. It means that the sequence will repeat itself when a value is reached a second time. (So you treat two cases: non repeating [1,2,3,4] and repeating [1,2,1,2], no hybrid cases like [1,2,1,4]). If there is no repetition you should return []. 


This kata is followed by two other cycle detection algorithms: 
Loyd's: http://www.codewars.com/kata/cycle-detection-floyds-tortoise-and-the-hare
Bret's: http://www.codewars.com/kata/cycle-detection-brents-tortoise-and-hare

I tried it in Python, but could not do it. Can you solve it?
targets: def cycle(sequence):
    memo = {}
    for i, item in enumerate(sequence):
        if item in memo:
            return [memo[item], i - memo[item]]
        memo[item] = i
    return []
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1204/D2:
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.

Kirk has a binary string $s$ (a string which consists of zeroes and ones) of length $n$ and he is asking you to find a binary string $t$ of the same length which satisfies the following conditions: For any $l$ and $r$ ($1 \leq l \leq r \leq n$) the length of the longest non-decreasing subsequence of the substring $s_{l}s_{l+1} \ldots s_{r}$ is equal to the length of the longest non-decreasing subsequence of the substring $t_{l}t_{l+1} \ldots t_{r}$; The number of zeroes in $t$ is the maximum possible.

A non-decreasing subsequence of a string $p$ is a sequence of indices $i_1, i_2, \ldots, i_k$ such that $i_1 < i_2 < \ldots < i_k$ and $p_{i_1} \leq p_{i_2} \leq \ldots \leq p_{i_k}$. The length of the subsequence is $k$.

If there are multiple substrings which satisfy the conditions, output any.


-----Input-----

The first line contains a binary string of length not more than $10^5$.


-----Output-----

Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.


-----Examples-----
Input
110

Output
010

Input
010

Output
010

Input
0001111

Output
0000000

Input
0111001100111011101000

Output
0011001100001011101000



-----Note-----

In the first example:  For the substrings of the length $1$ the length of the longest non-decreasing subsequnce is $1$;  For $l = 1, r = 2$ the longest non-decreasing subsequnce of the substring $s_{1}s_{2}$ is $11$ and the longest non-decreasing subsequnce of the substring $t_{1}t_{2}$ is $01$;  For $l = 1, r = 3$ the longest non-decreasing subsequnce of the substring $s_{1}s_{3}$ is $11$ and the longest non-decreasing subsequnce of the substring $t_{1}t_{3}$ is $00$;  For $l = 2, r = 3$ the longest non-decreasing subsequnce of the substring $s_{2}s_{3}$ is $1$ and the longest non-decreasing subsequnce of the substring $t_{2}t_{3}$...

I tried it in Python, but could not do it. Can you solve it?
targets: S = list(map(int,input().strip()))
N = len(S)

stack = []
for i in range(N):
    s = S[i]
    if s == 0 and stack and stack[-1][0] == 1:
        stack.pop()
    else:
        stack.append((s, i))
T = S[:]
if stack:
    for i in tuple(map(list, zip(*stack)))[1]:
        T[i] = 0

print(''.join(map(str, T)))
inputs: Solve in Python:
You are given an array $a$ consisting of $n$ integers. You have to find the length of the smallest (shortest) prefix of elements you need to erase from $a$ to make it a good array. Recall that the prefix of the array $a=[a_1, a_2, \dots, a_n]$ is a subarray consisting several first elements: the prefix of the array $a$ of length $k$ is the array $[a_1, a_2, \dots, a_k]$ ($0 \le k \le n$).

The array $b$ of length $m$ is called good, if you can obtain a non-decreasing array $c$ ($c_1 \le c_2 \le \dots \le c_{m}$) from it, repeating the following operation $m$ times (initially, $c$ is empty):  select either the first or the last element of $b$, remove it from $b$, and append it to the end of the array $c$. 

For example, if we do $4$ operations: take $b_1$, then $b_{m}$, then $b_{m-1}$ and at last $b_2$, then $b$ becomes $[b_3, b_4, \dots, b_{m-3}]$ and $c =[b_1, b_{m}, b_{m-1}, b_2]$.

Consider the following example: $b = [1, 2, 3, 4, 4, 2, 1]$. This array is good because we can obtain non-decreasing array $c$ from it by the following sequence of operations:  take the first element of $b$, so $b = [2, 3, 4, 4, 2, 1]$, $c = [1]$;  take the last element of $b$, so $b = [2, 3, 4, 4, 2]$, $c = [1, 1]$;  take the last element of $b$, so $b = [2, 3, 4, 4]$, $c = [1, 1, 2]$;  take the first element of $b$, so $b = [3, 4, 4]$, $c = [1, 1, 2, 2]$;  take the first element of $b$, so $b = [4, 4]$, $c = [1, 1, 2, 2, 3]$;  take the last element of $b$, so $b = [4]$, $c = [1, 1, 2, 2, 3, 4]$;  take the only element of $b$, so $b = []$, $c = [1, 1, 2, 2, 3, 4, 4]$ — $c$ is non-decreasing. 

Note that the array consisting of one element is good.

Print the length of the shortest prefix of $a$ to delete (erase), to make $a$ to be a good array. Note that the required length can be $0$.

You have to answer $t$ independent test cases.


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains...
targets: def solve():
    n = int(input())
    a = list(map(int, input().split()))
    prev = a[-1]
    cnt = 1
    status = 0
    for i in range(n - 2, -1, -1):
        v = a[i]
        if status == 0:
            if v < prev:
                status = 1
            cnt += 1
            prev = v
            continue
        elif status == 1:
            if v > prev:
                break
            cnt += 1
            prev = v
            continue
    print(n - cnt)

t = int(input())
for _ in range(t):
    solve()