TnT/CodeNet4Repair huggingface dataset:
Problem:
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.  The problem has N test cases.  For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively.  See the Output section for the output format.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
i = []

for x in range(n):
    x = input()
    i.append(x)    
print(i)
    
print('AC x ',i.count('AC'))
print('WA x ',i.count('WA'))
print('TLE x ',i.count('TLE'))
print('RE x ',i.count('RE'))

Fixed, accepted submission:
n = int(input())
i = []

for x in range(n):
    x = input()
    i.append(x)    

    
print('AC x',i.count('AC'))
print('WA x',i.count('WA'))
print('TLE x',i.count('TLE'))
print('RE x',i.count('RE'))
Problem:
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs.  Determine if it is possible that there are exactly X cats among these A + B animals.

Incorrect submission (verdict: Wrong Answer):
a,b,c=map(int, input().split())
print('YES' if a+b<=c<=a+b else 'NO')

Fixed, accepted submission:
a, b, x = map(int, input().split())
print('YES' if a <= x <= b+a else 'NO')
Problem:
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.  The person numbered i says:    * "In front of me, exactly A_i people are wearing hats with the same color as mine."  Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.  Since the count can be enormous, compute it modulo 1000000007.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
List=list(map(int,input().split()))

arr = [0,0,0]
ans = 1

for i in range(len(List)):
  a = List[i]
  
  b = arr.count(a)
  
  if b > 0:
    j = arr.index(a)
    arr[j] = a + 1
    
    ans = ans * b
    print(b)
    
print(ans)

Fixed, accepted submission:
n=int(input())
List=list(map(int,input().split()))

arr = [0,0,0]
ans = 1
MOD = 1000000007

for i in range(len(List)):
  a = List[i]
  
  b = arr.count(a)
  
  if b > 0:
    j = arr.index(a)
    arr[j] = a + 1
    
    ans = ans * b
    
  else:
    ans = 0
    break
    
print(ans%MOD)
Problem:
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows:    * Rating 1-399 : gray   * Rating 400-799 : brown   * Rating 800-1199 : green   * Rating 1200-1599 : cyan   * Rating 1600-1999 : blue   * Rating 2000-2399 : yellow   * Rating 2400-2799 : orange   * Rating 2800-3199 : red  Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.   Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.   Find the minimum and maximum possible numbers of different colors of the users.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
a=list(map(int,input().split()))
s=set()
fr=0
for i in a:
  x=i//400
  if(x<8):s.add(x)
  else:fr+=1
ki=len(s)
print(min(ki+fr,8))

Fixed, accepted submission:
n=int(input())
a=list(map(int,input().split()))
s=set()
fr=0
for i in a:
  x=i//400
  if(x<8):s.add(x)
  else:fr+=1
ki=len(s)
ma=ki+fr
mi=max(ki,1)
print(mi,ma)
Problem:
You are given three integers A, B and C.  Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

Incorrect submission (verdict: Wrong Answer):
hen=list(map(int,input().split()))
if max(hen)>sum(hen)-max(hen):
  print("Yes")
else:
  print("No")

Fixed, accepted submission:
hen=set(list(map(int,input().split())))
if len(hen)==1:
  print("Yes")
else:
  print("No")
Problem:
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym.  The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it.  Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up.  Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
cnt = 0

for i in range(n):
    a = int(input())
    if i ==0 and a==2:
        print(0)
        break
    else:
        if a == 2:
            print(i)
            break
        if a == 1:
            print(-1)
            break        

Fixed, accepted submission:
n = int(input())
a = [0]
for i in range(n):
    a.append(int(input()))
cnt = 0
a_before = 1
while cnt <= n:
    cnt += 1
    if a[a_before] == 2:
        print(cnt)
        break
    a_before = a[a_before]
else:
    print(-1)
Problem:
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Incorrect submission (verdict: Wrong Answer):
n = int(input())
l = []
for i in range(1,101):
    for j in range(1,11):
        k = i**j
        if k <=n:
            l.append(k)
print('{}'.format(max(l)))


Fixed, accepted submission:
n = int(input())
l = []
for i in range(1,101):
    for j in range(2,11):
        k = i**j
        if k <=n:
            l.append(k)
print('{}'.format(max(l)))
Problem:
宝物がたくさん収蔵されている博物館に、泥棒が大きな風呂敷を一つだけ持って忍び込みました。盗み出したいものはたくさんありますが、風呂敷が耐えられる重さが限られており、これを超えると風呂敷が破れてしまいます。そこで泥棒は、用意した風呂敷を破らず且つ最も価値が高くなるようなお宝の組み合わせを考えなくてはなりません。  風呂敷が耐えられる重さ W、および博物館にある個々のお宝の価値と重さを読み込んで、重さの総和が W を超えない範囲で価値の総和が最大になるときの、お宝の価値総和と重さの総和を出力するプログラムを作成してください。ただし、価値の総和が最大になる組み合わせが複数あるときは、重さの総和が小さいものを出力することとします。

Incorrect submission (verdict: Wrong Answer):
while True:
    cnt = 0
    cnt += 1
    maxweight = int(input())
    if maxweight == 0:
        break
    length = int(input())
    tresure = []
    dp = [[0 for n in range(length+1)] for m in range(maxweight + 1)]
    answeight = 0
    ansvalue = 0
    for l in range(length):
        v,w = (int(n) for n in input().split(","))
        tresure.append([v,w])
    for outer in range(length + 1):
        if outer == 0:
            continue
        weight = tresure[outer-1][1]
        value = tresure[outer-1][0]
        dp[weight][outer] = max(dp[weight][outer],value)
        for inner in range(maxweight + 1):
            if dp[inner][outer - 1] != 0:
                beforevalue = dp[inner][outer - 1]
                beforeweight = inner
                dp[inner][outer]  = max(beforevalue,dp[inner][outer])
                if beforeweight + weight <= maxweight:
                    nowvalue = dp[beforeweight + weight][outer]
                    dp[beforeweight + weight][outer] = max(nowvalue,beforevalue+value)
    #print(dp)
    for a in range(maxweight+1):
        #print(dp[a][length])
        if ansvalue < dp[a][length]:
            ansvalue = dp[a][length]
            answeight = a
    print("Case {0}:".format(cnt))
    print(ansvalue)
    print(answeight)

Fixed, accepted submission:
cnt = 0
while True:
    cnt += 1
    maxweight = int(input())
    if maxweight == 0:
        break
    length = int(input())
    tresure = []
    dp = [[0 for n in range(length+1)] for m in range(maxweight + 1)]
    answeight = 0
    ansvalue = 0
    for l in range(length):
        v,w = (int(n) for n in input().split(","))
        tresure.append([v,w])
    for outer in range(length + 1):
        if outer == 0:
            continue
        weight = tresure[outer-1][1]
        value = tresure[outer-1][0]
        dp[weight][outer] = max(dp[weight][outer],value)
        for inner in range(maxweight + 1):
            if dp[inner][outer - 1] != 0:
                beforevalue = dp[inner][outer - 1]
                beforeweight = inner
                dp[inner][outer]  = max(beforevalue,dp[inner][outer])
                if beforeweight + weight <= maxweight:
                    nowvalue = dp[beforeweight + weight][outer]
                    dp[beforeweight + weight][outer] = max(nowvalue,beforevalue+value)
    #print(dp)
    for a in range(maxweight+1):
        #print(dp[a][length])
        if ansvalue < dp[a][length]:
            ansvalue = dp[a][length]
            answeight = a
    print("Case {0}:".format(cnt))
    print(ansvalue)
    print(answeight)
Problem:
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A.  The problem has N test cases.  For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively.  See the Output section for the output format.

Incorrect submission (verdict: Wrong Answer):
n=int(input())
dic={}
dic["AC"]=0
dic["WA"]=0
dic["TLE"]=0
dic["RE"]=0
for x in range(n):
  a=input()
  dic[a]=dic[a]+1
print("AC",x,dic["AC"])
print("WA",x,dic["WA"])
print("TLE",x,dic["TLE"])
print("RE",x,dic["RE"])

Fixed, accepted submission:
n=int(input())
dic={}
dic["AC"]=0
dic["WA"]=0
dic["TLE"]=0
dic["RE"]=0
for x in range(n):
  a=input()
  dic[a]=dic[a]+1
print("AC","x",dic["AC"])
print("WA","x",dic["WA"])
print("TLE","x",dic["TLE"])
print("RE","x",dic["RE"])
Problem:
Write a program which solve a simultaneous equation:      ax + by = c   dx + ey = f      The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.

Incorrect submission (verdict: Wrong Answer):
a, b, c, d, e, f=map(float,input().split())
k=(a*e)-(b*d)
xval=(c*e)-(b*f)
yval=(a*f)-(c*d)
print("%.3f %.3f"%(xval/k,yval/k))

Fixed, accepted submission:
import sys
for line in sys.stdin.readlines():
    a, b, c, d, e, f=map(float,line.split())
    k=(a*e)-(b*d)
    xval=(c*e)-(b*f)
    yval=(a*f)-(c*d)
    g=xval/k
    h=yval/k
    if(xval==0):
        g=0
    if(yval==0):
        h=0
    print("%.3f %.3f"%(g,h))
Problem:
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}.  Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.

Incorrect submission (verdict: Wrong Answer):
x = list(input())
y = list(input())

if x[0] == y[2] and x[1] == y[1] and x[2] == y[0]:
    print('yes')
else:
    print('no')

Fixed, accepted submission:
x = list(input())
y = list(input())

if x[0] == y[2] and x[1] == y[1] and x[2] == y[0]:
    print('YES')
else:
    print('NO')
Problem:
We ask you to select some number of positive integers, and calculate the sum of them.  It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.  Your objective is to make the sum congruent to C modulo B. Determine whether this is possible.  If the objective is achievable, print `YES`. Otherwise, print `NO`.

Incorrect submission (verdict: Wrong Answer):
import sys
A,B,C = map(int,input().split())
for i in range(1,B+1):
    if A * i % B == C:
        print("Yes")
        sys.exit()
print("No")

Fixed, accepted submission:
import sys
A,B,C = map(int,input().split())
for i in range(1,B+1):
    if A * i % B == C:
        print("YES")
        sys.exit()
print("NO")
Problem:
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b.  Determine whether the arrangement of the poles is beautiful.

Incorrect submission (verdict: Wrong Answer):
s=input()
print(s[0]+str((len(s)-2))+s[-1])

Fixed, accepted submission:
a,b,c=map(int,input().split())
if b-a==c-b:
  print("YES")
else:
  print("NO")
Problem:
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.  The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.  How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?

Incorrect submission (verdict: Wrong Answer):
N = int(input())
H = int(input())
W = int(input())
print((H-N+1)*(W-N+1))

Fixed, accepted submission:
N = int(input())
H = int(input())
W = int(input())
print((N-H+1)*(N-W+1))
Problem:
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows:    * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there.   * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1.  You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.

Incorrect submission (verdict: Wrong Answer):
N, M = map(int, input().split())

for i in range(M):
  print(i + 1, N - i)
  
  

Fixed, accepted submission:
N, M = map(int, input().split())

x = int(M // 2)

if M % 2 == 0:
  for i in range(x):
    print(i + 1, M + 1 - i)
    print(M + 2 + i, 2 * M + 1 - i)  
    
else:
  for i in range(x):
    print(i + 1, M - i)
  for j in range(x + 1):
    print(M + 1 + j, 2 * M + 1 - j)
Problem:
E869120 found a chest which is likely to contain treasure.   However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters.   He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`.  One more thing he found is a sheet of paper with the following facts written on it:    * Condition 1: The string S contains a string T as a contiguous substring.   * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1.  Print the string S.   If such a string does not exist, print `UNRESTORABLE`.

Incorrect submission (verdict: Wrong Answer):
s = list(input())
t = list(input())
ans = []
ok = -1
f = -1
for i in range(len(s)-len(t)+1):
    # print(s[i])
    a = 0
    if s[i] == '?' or s[i] == t[0]:
        if f is -1:
            f=i
        for j in range(len(t)):
            # print(i,j,s[i+j],t[j])
            if s[i+j] != '?' and s[i+j] != t[j]:
                break
            a = j
        if ok == -1 and a == len(t)-1:
            ok = i
if ok == -1:
    print('UNRESTORABLE')
    exit()
else:
    for i in range(len(s)):
        if s[i] == '?':
            s[i] = 'a'
        if  ok<= i <= ok+len(t)-1:
            s[i] = t[i-ok]

print(s)


Fixed, accepted submission:
s = list(input())
t = list(input())
ans = []
ok = -1
f = -1
if len(s)<len(t):
    print('UNRESTORABLE')
    exit()

for i in range(len(s)-len(t)+1):
    # print(s[i])
    a = 0
    if s[i] == '?' or s[i] == t[0]:
        if f is -1:
            f=i
        for j in range(len(t)):
            # print(i,j,s[i+j],t[j])
            if s[i+j] != '?' and s[i+j] != t[j]:
                break
            a = j
        if a == len(t)-1:
            ok = i
if ok == -1:
    print('UNRESTORABLE')
    exit()
else:
    for i in range(len(s)):
        if s[i] == '?':
            s[i] = 'a'
        if  ok<= i <= ok+len(t)-1:
            s[i] = t[i-ok]

print(''.join(s))
Problem:
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.

Incorrect submission (verdict: Wrong Answer):
a , b = map(int, input().split())
c = a*b
if c%2 == 1:
  print("odd")
else:
  print("even")

Fixed, accepted submission:
a , b = map(int, input().split())
c = a*b
if c%2 == 1:
  print("Odd")
else:
  print("Even")
Problem:
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters.  Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way.  You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.

Incorrect submission (verdict: Wrong Answer):
s=input()
for i in range(len(s)):
    for j in range(i,len(s)):
        if s[i]=="C":
            if s[j]=="F":
                print("Yes")
                break
else:
    print("No")

Fixed, accepted submission:
s=input()
a="No"
for i in range(len(s)):
    for j in range(i,len(s)):
        if s[i]=="C":
            if s[j]=="F":
                a="Yes"
                break
print(a)
Problem:
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.

Incorrect submission (verdict: Wrong Answer):
a,b,c = map(int,input().split())

if a <= b <= c:
    print("Yes")
else:
    print("No")




Fixed, accepted submission:
a,b,c = map(int,input().split())

if a <= c and b>=c:
    print("Yes")
else:
    print("No")
Problem:
In this problem, we only consider strings consisting of lowercase English letters.  Strings s and t are said to be **isomorphic** when the following conditions are satisfied:    * |s| = |t| holds.   * For every pair i, j, one of the following holds:     * s_i = s_j and t_i = t_j.     * s_i \neq s_j and t_i \neq t_j.  For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not.  A string s is said to be in **normal form** when the following condition is satisfied:    * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison.  For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`.  You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.

Incorrect submission (verdict: Wrong Answer):
N=int(input())


def combi(N,K):
    ans1=1
    ans2=1
    for i in range(K):
        ans1=ans1*(N-i)
        ans2=ans2*(K-i)
    ans=ans1//ans2
    return ans

def rec(n):
    if n==1:
        return 1
    else:
        ans=0
        for i in range(1,n+1):
            if i==1:
                ans+=1
            elif i<n:
                ans+=combi(n-1,n-i)
            elif i==n:
                ans+=rec(n-1)
        return ans

a=rec(N)

print(a)

Fixed, accepted submission:
N=int(input())

if N==1:
    print("a")
    exit()

from collections import deque
from collections import Counter
abc="abcdefghijklmnopqrstuvwxyz"
ans=["a"]

for i in range(1,N):
    d=deque(ans)
    ans=[]
    while d:
        temp=d.popleft()
        temp2=list(temp)
        cnt=Counter(temp2)
        L=len(cnt)
        for j in range(L+1):
            ans.append(temp+abc[j])

for a in ans:
    print(a)
Problem:
Snuke is giving cookies to his three goats.  He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).  Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Incorrect submission (verdict: Wrong Answer):
A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A + B) % 3 == 0:
    print("Yes")
else:
    print("No")

Fixed, accepted submission:
A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A + B) % 3 == 0:
    print("Possible")
else:
    print("Impossible")
Problem:
Jerry is a little mouse. He is trying to survive from the cat Tom. Jerry is carrying a parallelepiped-like piece of cheese of size _A_ × _B_ × _C_. It is necessary to trail this cheese to the Jerry's house. There are several entrances in the Jerry's house. Each entrance is a rounded hole having its own radius _R_. Could you help Jerry to find suitable holes to be survive?  Your task is to create a program which estimates whether Jerry can trail the cheese via each hole. The program should print "OK" if Jerry can trail the cheese via the corresponding hole (without touching it). Otherwise the program should print "NA".  You may assume that the number of holes is less than 10000.

Incorrect submission (verdict: Wrong Answer):
import math
while True:
	d,w,h = map(int,input().split())
	if d == 0:
		break
	n = int(input())
	dist = [d**2+w**2, d**2+h**2, w**2+h**2]
	leng = min(dist)
	for i in range(n):
		if leng-int(input()) > 0:
			print('OK')
		else:
			print('NA')

Fixed, accepted submission:
import math
while True:
	d,w,h = map(int,input().split())
	if d == 0:
		break
	n = int(input())
	dist = [math.sqrt(d**2+w**2), math.sqrt(d**2+h**2), math.sqrt(w**2+h**2)]
	leng = min(dist)
	for i in range(n):
		if 2*int(input())-leng > 0:
			print('OK')
		else:
			print('NA')
Problem:
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.

Incorrect submission (verdict: Wrong Answer):
import math

N = int(input())

class remainder():
    def __init__(self, mod=(10**9 + 7)):
        self.mod = mod
    
    def mul(self, a, b):
        return ((a % self.mod) * (b % self.mod)) % self.mod

    def pow(self, a, b):
        bp = []
        #bp.append(1)
        bp.append(a)

        n = len(bin(b)) - 2
        for x in range(n):
            bp.append(bp[x]**2 % self.mod)

        res = 1
        for x in range(n):
            if b >> x & 1:
                res = (res * bp[x]) % self.mod
        
        return res

    def div(self, a, b):
        
        return self.mul(a, self.pow(b, self.mod - 2))


def findAllOfPrime(N):
    prime = [True] * (N + 1)
    prime[0] = False
    prime[1] = False

    x = 2
    while x**2 <= N:
        if prime[x]:
            j = 2
            while x * j <= N:
                prime[x * j] = False
                j += 1
        x += 1

    return prime

re = remainder()
primes = set()
for n in range(1, N + 1):
    prime = findAllOfPrime(n)
    for i, p in enumerate(prime):
        if p:
            primes.add(i)

print(primes)

A = math.factorial(N)
ans = 1
for p in primes:
    cnt = 1
    while A % p == 0:
        cnt += 1
        A //= p
    ans = re.mul(ans, cnt)
print(ans)




Fixed, accepted submission:
import math

N = int(input())

class remainder():
    def __init__(self, mod=(10**9 + 7)):
        self.mod = mod
    
    def mul(self, a, b):
        return ((a % self.mod) * (b % self.mod)) % self.mod

    def pow(self, a, b):
        bp = []
        #bp.append(1)
        bp.append(a)

        n = len(bin(b)) - 2
        for x in range(n):
            bp.append(bp[x]**2 % self.mod)

        res = 1
        for x in range(n):
            if b >> x & 1:
                res = (res * bp[x]) % self.mod
        
        return res

    def div(self, a, b):
        
        return self.mul(a, self.pow(b, self.mod - 2))


def findAllOfPrime(N):
    prime = [True] * (N + 1)
    prime[0] = False
    prime[1] = False

    x = 2
    while x**2 <= N:
        if prime[x]:
            j = 2
            while x * j <= N:
                prime[x * j] = False
                j += 1
        x += 1

    return prime

re = remainder()
primes = set()
for n in range(1, N + 1):
    prime = findAllOfPrime(n)
    for i, p in enumerate(prime):
        if p:
            primes.add(i)

A = math.factorial(N)
ans = 1
for p in primes:
    cnt = 1
    while A % p == 0:
        cnt += 1
        A //= p
    ans = re.mul(ans, cnt)
print(ans)