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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>


using namespace std;


int comb2(int x) {
    if (x < 2)
        return 0;
    return (x * (x - 1)) / 2;
}


int main() {
    int N, P;
    string S;
    cin >> N >> P;
    cin >> S;

    if (P == 2 || P == 5) {
        int ans = 0;
        for (int i = S.length() - 1; i >= 0; i--) {
            if ((S[i] - '0') % P == 0) {
                ans += (i + 1);
            }
        }
        cout << ans << endl;
    }

    vector<int> r_cnt(P, 0);
    r_cnt[0]++;
    int r = 0;
    int base = 1;
    for (int i = S.length() - 1; i >= 0; i--) {
        r += (S[i] - '0') * base;
        r %= P;
        r_cnt[r]++;
        base = (base * 10) % P;
    }

    int ans = 0;
    for (auto c : r_cnt) {
        ans += comb2(c);
    }
    cout << ans << endl;
}