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;

const int N = 2e5 + 5;
int n, p;
long long pre[N], cnt[N], po10[N];
string s;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    cin >> n >> p >> s;
    po10[0] = 1;
    for(int i = 1; i <= n; i++){
        po10[i] = (po10[i - 1] * 10) % p;
    }
    reverse(s.begin(), s.end());
    if(p == 2){
        long long ans = 0;
        for(int i = 0; i < n; i++){
            if(s[i] == '0' || s[i] == '2'){
                ans += (long long) n - i;
            }
        }
        cout << ans << '\n';
    } else if(p == 5){
long long ans = 0;
        for(int i = 0; i < n; i++){
            if(s[i] == '0' || s[i] == '5'){
                ans += (long long) n - i;
            }
        }
        cout << ans << '\n';
    } else {
        long long ans = 0;
        long long las = 0;
        for(int i = 0; i < n; i++){
                int k = i + 1;
                long long he = (pre[k - 1] % p + ((s[i] - '0') * po10[i]) % p) % p;
                pre[k] = he;
        }
        cnt[0] = 1;
        for(int i = 1; i <= n; i++){
            ans += cnt[pre[i]];
            cnt[pre[i]]++;
        }
        cout << ans << '\n';
    }
    return 0;
}