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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/**********************************************************************
File    .cpp
Author  Takahiro Yamazaki
**********************************************************************/
/**********************************************************************
Include Libraries
**********************************************************************/
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, a, b) for (int i = a; i < b; i++)
#define each(i, mp) for(auto& i:mp)
/**********************************************************************
Declare variables
**********************************************************************/
/**********************************************************************
Declare instance
**********************************************************************/
/**********************************************************************
Proto_type_Declare functions
**********************************************************************/

/**********************************************************************
Main
**********************************************************************/
int main() {
  ll ans = 0;
  ll n,p;
  cin >> n >> p;
  string s;
  cin >> s;
  reverse(s.begin(), s.end());

  map<ll, ll> mp;
  mp[0] = 1;
  ll now = 0, sum = 0, ten = 1,a,b;
  if (p == 2 || p == 5) {
    each(e,s){
      if ((e - '0')%p == 0) {
        mp[0]++;
      }
    }
  }
  else {
    each(e, s) {
      b = ((e - '0') * ten) % p;
      now = (now + b) % p;
      ten *= 10;
      ten %= p;
      mp[now]++;
    }
  }
  each(e, mp) {
    ans += e.second * (e.second - 1) / 2;
  }
  cout << ans << '\n';
  return 0;

}
/**********************************************************************
Functions
**********************************************************************/