Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import math
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9 +7
INF = float('inf')

#############
# Functions #
#############

kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n):
    return kaijo_memo[n]
  if(len(kaijo_memo) == 0):
    kaijo_memo.append(1)
  while(len(kaijo_memo) <= n):
    kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n):
    return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0):
    gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n):
    gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if(n == r):
    return 1
  if(n < r or r < 0):
    return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])
    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])
    return arr

#############
# Main Code #
#############

N,P = map(int,input().split())
S = list(input())

mods = []
count = 0

for digit in S:
  print(digit)
  mods.append(int(digit))
  if int(digit)%P == 0:
    count += 1
  if len(mods) > 1:
    for i in range(len(mods)-1):
      mods[i] = (mods[i]*10 + int(digit)) % P
      if mods[i] == 0:
        count += 1

print(count)