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: Accepted)

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n, p;
    string s;
    cin >> n >> p >> s;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        a[i] = s[i] - '0';
    long long ans = 0;
    if(10 % p == 0) {
        for(int i = 0 ; i < n ; i++)
            if(a[i] % p == 0)
                ans += i + 1;
        cout << ans << '\n';
        return 0;
    }
    reverse(a.begin(), a.end());
    vector<int> csum(n + 1, 0);
    int pw = 1;
    for(int i = 0; i < n; i++) {
        csum[i + 1] += (csum[i] + a[i] * pw % p) % p;
        pw = (pw * 10) % p;
    }
    
    // map<int, int> mp;
    // for(int i = 0; i < n + 1; i++)
    //     mp[csum[i]]++;
    // for(auto &iter : mp)
    //     ans += iter.second * (iter.second - 1) /2;
    vector<int> cnt(p);
    for(int i = 0; i < n + 1; i++) {
        ans += cnt[csum[i]];
        cnt[csum[i]]++;
    }
    cout << ans << '\n';

    return 0;
}