TnT/CodeNet4Repair huggingface dataset:
Problem:
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times:    * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n.  What is the largest possible sum of the integers written on the blackboard after K operations?

Incorrect submission (verdict: Wrong Answer):
def main():
    A,B,C = list(map(int, input().split()))
    K = int(input())

    return A + B + C + max(A,B,C) **K - max(A,B,C)

if __name__ == '__main__':
    print(main())


Fixed, accepted submission:
def main():
    A,B,C = list(map(int, input().split()))
    K = int(input())

    for i in range(K):
        if A == max(A,B,C):
            A = A*2
        elif B == max(A,B,C):
            B = B*2
        else:
            C = C*2
    return A+B+C

if __name__ == '__main__':
    print(main())
Problem:
Takahashi loves numbers divisible by 2.  You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.  Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.  For example,    * 6 can be divided by 2 once: 6 -> 3.   * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.   * 3 can be divided by 2 zero times.

Incorrect submission (verdict: Wrong Answer):
A = int(input())
s = 2
for i in range(1,A):
  if A == 1:
    print(1)
  elif s <= A:
    s = 2**i
  else:
    print(s//2)
  break

Fixed, accepted submission:
A = int(input())
s = 2
for i in range(1,A):
  if A == 1:
    print(1)
    break;
  elif A == 2:
    print(2)
    break;
  elif s <= A:
    s = 2**i
  else:
    break;
      
print(s//2)
Problem:
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives.  You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative.  For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins.  If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Incorrect submission (verdict: Wrong Answer):
n = int(input())
a = [input().split() for i in range(n)]
yen = 0
bit = 0
for f in a:
  if f[1] == "JPY":
    yen += int(f[0])
  else:
    bit += float(f[0])
print(yen + int(380000 * bit))    

Fixed, accepted submission:
n = int(input())
xu = [input().split() for i in range(n)]
y = 0
b = 0
for j in xu:
  if j[1] == "JPY":
    y += int(j[0])
  else:
    b += float(j[0])
ans = y + 380000 * b
print(ans)
Problem:
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.  Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Incorrect submission (verdict: Wrong Answer):
a,b,c,x,y=map(int,input().split())
if a+b<=2*c: print(a*x+b*y)
else:
    ans=min(x,y)*2*c
    print(ans)
    dif=abs(x-y)
    if x>y:
        if a>=2*c:
            ans+=2*dif*c
        else:
            ans+=a*dif
    else:
        if b>=2*c:
            ans+=2*dif*c
        else:
            ans+=b*dif
    print(ans)

Fixed, accepted submission:
a,b,c,x,y=map(int,input().split())
if a+b<=2*c: print(a*x+b*y)
else:
    ans=min(x,y)*2*c
    dif=abs(x-y)
    if x>y:
        if a>=2*c:
            ans+=2*dif*c
        else:
            ans+=a*dif
    else:
        if b>=2*c:
            ans+=2*dif*c
        else:
            ans+=b*dif
    print(ans)
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())
print("Even" if a*b%2 else "Odd")

Fixed, accepted submission:
a, b = map(int, input().split())
print("Odd" if a*b%2 else "Even")
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
import sys
df = list(map(int, list(str(input()))))
print(df)
counter = 0
for i in range(2**3):
    bin_i = format(i, '03b')
    counter += df[0]
    for j in range(3):
        print(j, counter)
        if bin_i[j] == '1':
            counter -= df[j+1]
        else:
            counter += df[j+1]
    if counter == 7:
        result = str(df[0])
        for j in range(3):
            if bin_i[j] =='0':
                result += '+' + str(df[j+1])
            else:
                result += '-' + str(df[j+1])
        result += '=7'
        print(result)
        sys.exit()
    else:
        print(counter)
        counter = 0

Fixed, accepted submission:
import sys
df = list(map(int, list(str(input()))))
counter = 0
for i in range(2**3):
    bin_i = format(i, '03b')
    counter += df[0]
    for j in range(3):
        if bin_i[j] == '1':
            counter -= df[j+1]
        else:
            counter += df[j+1]
    if counter == 7:
        result = str(df[0])
        for j in range(3):
            if bin_i[j] =='0':
                result += '+' + str(df[j+1])
            else:
                result += '-' + str(df[j+1])
        result += '=7'
        print(result)
        sys.exit()
    else:
        counter = 0
Problem:
We will buy a product for N yen (the currency of Japan) at a shop.  If we use only 1000-yen bills to pay the price, how much change will we receive?  Assume we use the minimum number of bills required.

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

print(n % 1000)

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

if n % 1000 == 0:
    print(0)
else:
    print(1000 - n % 1000)
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):
n,m,x,*l=map(int,open(0).read().split())
s=sum(i in l for i in range(x))
print(min(s,m-s))

Fixed, accepted submission:
a,b,x=map(int,input().split());print("NYOE S"[a<=x<=a+b::2])
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())
print('YES' if A<=C<=B else 'NO')

Fixed, accepted submission:
A,B,C = map(int,input().split())
print('Yes' if A<=C<=B else 'No')
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
import itertools
status = [(0, 1) for _ in range(3)]
state = list(itertools.product(*status))
num = [int(i) for i in input()]
for ptn in state:
  ans = num[0]
  ans_str = str(num[0])
  for i,ope in enumerate(ptn):
    if ope == 0:
      ans += num[i+1]
      ans_str += ("+" + str(num[i+1]))
    else:
      ans -= num[i+1]
      ans_str += ("-" + str(num[i+1]))
  if ans == 7:
    break
print(ans_str)


Fixed, accepted submission:
import itertools
status = [(0, 1) for _ in range(3)]
state = list(itertools.product(*status))
num = [int(i) for i in input()]
for ptn in state:
  ans = num[0]
  ans_str = str(num[0])
  for i,ope in enumerate(ptn):
    if ope == 0:
      ans += num[i+1]
      ans_str += ("+" + str(num[i+1]))
    else:
      ans -= num[i+1]
      ans_str += ("-" + str(num[i+1]))
  if ans == 7:
    break
print(ans_str+"=7")
Problem:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Incorrect submission (verdict: Wrong Answer):
a=input()
b=""

for i in a:
    if i=="0":
        b+="0"
    if i=="1":
        b+="1"
    if i=="B":
        b=b[:-1]

print(b)

Fixed, accepted submission:
a=int(input())
answer=0
for i in range(1,a+1):
  answer+=i
print(answer)
Problem:
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.  Integers M_1, D_1, M_2, and D_2 will be given as input.   It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.   Determine whether the date 2019-M_1-D_1 is the last day of a month.

Incorrect submission (verdict: Wrong Answer):
str  = [input() for i in range(2)]
s1 = str[0].split()
s2 = str[1].split()

if s1[0] == s2[1]:
  print("0")
else:
  print("1")


Fixed, accepted submission:
str  = [input() for i in range(2)]
s1 = str[0].split()
s2 = str[1].split()
 
if s1[0] != s2[0] and int(s1[1]) >= int(s2[1]):
  print("1")
else:
  print("0")
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())
s = []

ac = 0
wa = 0
tle = 0
re = 0

for i in range(n):
  s.append(input())
  if s[i] == "AC":
    ac += 1
  elif s[i] == "WA":
    wa += 1
  elif s[i] == "TLE":
    tle += 1
  else:
    re += 1

print("AC × " + str(ac))
print("WA × " + str(wa))
print("TLE × " + str(tle))
print("RE × "+ str(re))

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

ac = 0
wa = 0
tle = 0
re = 0

for i in range(n):
  s.append(input())
  if s[i] == "AC":
    ac += 1
  elif s[i] == "WA":
    wa += 1
  elif s[i] == "TLE":
    tle += 1
  else:
    re += 1

print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x "+ str(re))
Problem:
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i.  Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.

Incorrect submission (verdict: Wrong Answer):
S = list(input().split())
S = list(map(int, S))
print(sum(S))


Fixed, accepted submission:
S = list(input())
S = list(map(int, S))
print(sum(S))
Problem:
アイヅ放送協会の教育番組(ＡＨＫ教育)では、子ども向けの工作番組「あそんでつくろ」を放送しています。今回は棒で箱を作る回ですが、用意した１２本の棒を使って直方体ができるかを確かめたいと思います。ただし、棒は切ったり折ったりしてはいけません。      １２本の棒の長さが与えられるので、それらすべてを辺とする直方体が作れるかどうか判定するプログラムを作成せよ。

Incorrect submission (verdict: Wrong Answer):
li = list(map(int,input().split()))
so = sorted(li)
a = so[0]
b = so[4]
c = so[8]
res = False
for n in so[:4]:
    res = a==n
for n in so[4:8]:
    res = b==n
for n in so[8:]:
    res = c==n
print(res)


Fixed, accepted submission:
li = list(map(int,input().split()))
so = sorted(li)
a = so[0]
b = so[4]
c = so[8]
for n in so[:4]:
    res1=  a==n
for n in so[4:8]:
    res2=  b==n
for n in so[8:]:
    res3= c==n
res = 'yes' if res1 and res3 and res2 else 'no'
print(res)
Problem:
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters.  Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s.  Given the name of the contest, print the abbreviation of the name.

Incorrect submission (verdict: Wrong Answer):
s=input()
print('A'+s[1]+'C')

Fixed, accepted submission:
s=input().split()[1]
print('A'+s[0]+'C')
Problem:
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b.  The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.

Incorrect submission (verdict: Wrong Answer):
while 1:
    a,op,b = input().split()
    a = int(a)
    b = int(b)

    if op == '?' :
        break
    if op == '+' :
        print(a + b)
    if op == '-' :
        print(a - b)
    if op == '*' :
        print(a * b)
    if op == '/' :
        print(a / b)

Fixed, accepted submission:
while 1:
    a,op,b = input().split()
    a = int(a)
    b = int(b)

    if op == '?' :
        break
    if op == '+' :
        print(a + b)
    if op == '-' :
        print(a - b)
    if op == '*' :
        print(a * b)
    if op == '/' :
        print(a // b)
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):
A, B, C = map(int, input().split(" "))

for i in range(1, 1000):
  if A * i % B == C:
    print("Yes")
    exit(0)
print("No")

Fixed, accepted submission:
A, B, C = map(int, input().split(" "))
l = []
for i in range(10000000):
  tmp = A*i%B
  if tmp in l:
    break
  l.append(tmp)
if C in l:
  print("YES")
else:
  print("NO")
Problem:
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.  Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.

Incorrect submission (verdict: Wrong Answer):
a,b,c,x,y = list(map(int,input().split()))
cost = []
tmp = 0
i = 0

while i <= max(x,y) :
    tmp += i * 2 * c
    if i >= x :
        tmp += (y-i) * b
    elif i >= y:
        tmp += (x-i) * a
    else:
        tmp += (x - i) * a + (y - i) * b
    cost.append(tmp)
    i += 1
    tmp = 0

print(min(cost))
    


Fixed, accepted submission:
a,b,c,x,y = list(map(int,input().split()))
cost = []
tmp = 0
i = 0

while i <= max(x,y) :
    tmp += i * 2 * c
    tmp += max(x-i,0)*a + max(y-i,0) * b
    cost.append(tmp)
    i += 1
    tmp = 0

print(min(cost))
Problem:
Sitting in a station waiting room, Joisino is gazing at her train ticket.  The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).  In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds.  The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Incorrect submission (verdict: Wrong Answer):
input_str = input()
A = int(input_str[0])
B = int(input_str[1])
C = int(input_str[2])
D = int(input_str[3])

list = []
list.append(A + B + C + D)
list.append(A + B + C - D)
list.append(A + B - C + D)
list.append(A + B - C - D)
list.append(A - B + C + D)
list.append(A - B + C - D)
list.append(A - B - C + D)
list.append(A - B - C - D)

num = list.index(7)
print(num)

if num <= 3:
    op1 = '+'
else:
    op1 = '-'

if num == 0 or num == 1 or num == 4 or num == 5:
    op2 = '+'
else:
    op2 = '-'

if num % 2 == 0:
    op3 = '+'
else:
    op3 = '-'

print('{}{}{}{}{}{}{}=7'.format(A,op1,B,op2,C,op3,D))


Fixed, accepted submission:
input_str = input()
A = int(input_str[0])
B = int(input_str[1])
C = int(input_str[2])
D = int(input_str[3])

list = []
list.append(A + B + C + D)
list.append(A + B + C - D)
list.append(A + B - C + D)
list.append(A + B - C - D)
list.append(A - B + C + D)
list.append(A - B + C - D)
list.append(A - B - C + D)
list.append(A - B - C - D)

num = list.index(7)

if num <= 3:
    op1 = '+'
else:
    op1 = '-'

if num == 0 or num == 1 or num == 4 or num == 5:
    op2 = '+'
else:
    op2 = '-'

if num % 2 == 0:
    op3 = '+'
else:
    op3 = '-'

print('{}{}{}{}{}{}{}=7'.format(A,op1,B,op2,C,op3,D))
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
import sys
s = input()
i = 0
while s[i:5] != '':
    if s[i:5] == 'dream':
        if s[i+5:i+10] == 'dream':
            i += 5
        elif s[i+5:i+10] == 'erase':
            i += 5
        elif s[i+5:i+7] == 'er':
            i += 7
        else:
            print('No')
            sys.exit(0)
    elif s[i:5] == 'erase':
        if s[i+5:i+10] == 'dream':
            i += 5
        elif s[i+5:i+10] == 'erase':
            i += 5
        elif s[i+5:i+6] == 'r':
            i += 6
        else:
            print('No')
            sys.exit(0)
    else:
        print('No')
        sys.exit(0)

print('Yes')

Fixed, accepted submission:
import sys
s = input()
i = 0
while s[i:i+5] != '':
    if s[i:i+5] == 'dream':
        if s[i+5:i+10] == 'dream':
            i += 5
        elif s[i+5:i+10] == 'erase':
            i += 5
        elif s[i+5:i+7] == 'er':
            i += 7
        else:
            i += 5
    elif s[i:i+5] == 'erase':
        if s[i+5:i+10] == 'dream':
            i += 5
        elif s[i+5:i+10] == 'erase':
            i += 5
        elif s[i+5:i+6] == 'r':
            i += 6
        else:
            i += 5
    else:
        print('NO')
        sys.exit(0)
print('YES')
Problem:
Write a program which reads two integers a and b, and calculates the following values:    * a ÷ b: d (in integer)   * remainder of a ÷ b: r (in integer)   * a ÷ b: f (in real number)

Incorrect submission (verdict: Wrong Answer):
a, b = map(int, input().split())
print("{} {} {}".format(a // b, a % b, float(a / b)))


Fixed, accepted submission:
a, b = map(int, input().split())
print("{} {} {:.5f}".format(a // b, a % b, float(a / b)))
Problem:
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.   According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.   Determine if he is correct.

Incorrect submission (verdict: Wrong Answer):
def main():
    c=[list(map(int,input().split())) for i in range(3)]
    a=[0,0,0]
    b=[0,0,0]
    for i in range(1,101):
        a[0]=i
        a[1]=-1
        a[2]=-1
        b[0]=c[0][0]-a[0]
        b[1]=c[0][1]-a[0]
        b[2]=c[0][2]-a[0]
        for j in range(1,3):
            if c[j][0]-b[0]!=c[j][1]-b[1] or c[j][1]-b[1]!=c[j][2]-b[2]:
                break
            a[j]=c[j][0]-b[0]
            if j == 2:
                print('YES')
                return
    print('NO')

if __name__=='__main__':
    main()


Fixed, accepted submission:
def main():
    c=[list(map(int,input().split())) for i in range(3)]
    a=[0,0,0]
    b=[0,0,0]
    for i in range(0,101):
        a[0]=i
        a[1]=-1
        a[2]=-1
        b[0]=c[0][0]-a[0]
        b[1]=c[0][1]-a[0]
        b[2]=c[0][2]-a[0]
        for j in range(1,3):
            if c[j][0]-b[0]!=c[j][1]-b[1] or c[j][1]-b[1]!=c[j][2]-b[2]:
                break
            a[j]=c[j][0]-b[0]

            if j == 2 and check(a,b):
                print('Yes')
                return
    print('No')

def check(a,b):
    for i in range(0,3):
        if a[i] >= 0 and b[i] >= 0 :
            continue
        else:
            return False
    return True

if __name__=='__main__':
    main()
Problem:
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times:    * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.

Incorrect submission (verdict: Wrong Answer):
s=input()[::-1]
a="dream"[::-1]
b="dreamer"[::-1]
c="erase"[::-1]
d="eraser"[::-1]
i=0
ans="Yes"
while i<len(s):
    if a in s[i:i+5]:
        i+=5
    elif b in s[i:i+7]:
        i+=7
    elif c in s[i:i+5]:
        i+=5
    elif d in s[i:i+6]:
        i+=6
    else :
        ans="No"
        break

print(ans)

Fixed, accepted submission:
s=input()[::-1]
a="dream"[::-1]
b="dreamer"[::-1]
c="erase"[::-1]
d="eraser"[::-1]
i=0
ans="YES"
while i<len(s):
    if a in s[i:i+5]:
        i+=5
    elif b in s[i:i+7]:
        i+=7
    elif c in s[i:i+5]:
        i+=5
    elif d in s[i:i+6]:
        i+=6
    else :
        ans="NO"
        break

print(ans)
Problem:
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.  Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.

Incorrect submission (verdict: Wrong Answer):
def colorful_subsequence(N: int, S: str)->int:
    MOD = (10 ** 9) + 7
    cum = {}
    for c in S:
        cum.setdefault(c, 0)
        cum[c] += 1

    values = list(cum.values())
    count = 0
    # for i, v in enumerate(values):
    for v in values:
            # temp = v
            # for v in values[i+1:]:
            #     temp = (temp * (1 + v)) % MOD
            # count = (count + temp) % MOD
        count *= (1 + v)
        count %= MOD

    return count


if __name__ == "__main__":
    N = int(input())
    S = input()
    ans = colorful_subsequence(N, S)
    print(ans)


Fixed, accepted submission:
def colorful_subsequence(N: int, S: str)->int:
    MOD = (10 ** 9) + 7
    cum = {}
    for c in S:
        cum.setdefault(c, 0)
        cum[c] += 1

    values = list(cum.values())
    count = 1
    for v in values:
        count *= (1 + v)
        count %= MOD

    return (count-1) % MOD


if __name__ == "__main__":
    N = int(input())
    S = input()
    ans = colorful_subsequence(N, S)
    print(ans)