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)

#include <bits/stdc++.h>
using namespace std;
int main() {
    long long N, P; cin >> N >> P;
    string s; cin >> s;
    vector<int> x(P, 0);
    long long d = 1;
    long long k = 0;
    x[k]++;
    long long ans = 0;
    for (int i = 1; i <= N; i++)
    {
        if(P == 2 || P == 5) {
            if((s[N-i] - '0') % P == 0) ans += N-i+1;
        }
        else {
            k = (k + (s[N-i] - '0') * d) % P;
            d = (d * 10) % P;
            x[k]++;
        }
    }

    if(P == 2 || P == 5) {
        cout << ans << endl;
    }
    else {
        ans = 0;
        for (int i = 0; i < P; i++)
        {
            ans += x[i] * (x[i]-1) / 2;
        }
        cout << ans << endl;
    }
}