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 <set>
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define LL long long
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define fast_io ios::sync_with_stdio(false)

const int maxn = 2e5 + 7;

char s[maxn];
LL vis[maxn];
deque<char>dq;

LL qpow(LL a,LL b, LL p) {
    LL res = 1;
    while(b) {
        if(b & 1LL) res = (res * a) % p;
        a = (a * a) % p;
        b >>= 1LL;
    }
    return res;
}

int main() {
    int n,p;
    while(~scanf("%d%d",&n,&p)) {
        scanf("%s",s);
        LL ans = 0;
        if(p == 2 || p == 5) {
            for(int i = 0 ; i < n ; i++) {
                if((s[i] - '0') % p == 0) ans += 1LL * (i + 1);
            }
        } else {
            LL sum = 0;
            memset(vis,0,sizeof(vis));
            vis[0] = 1;
            for(int i = 0 ; i < n ; i++) {
                sum = (sum * 10 + (s[i] - '0')) % p;
                LL val = (qpow(10,n-i-1,p) * sum) % p;;
                vis[val]++;
            }
            for(int i = 0 ; i < n ; i++) {
                ans += (vis[i] * (vis[i] - 1)) / 2;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}