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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
int N,S;
ll K;
int ans=0;
int A[10000];
int B [2000001];




int main(){
    cin >> N>>S;
    
    if(S == 2 || S==5){
        for(int i=0;i<N;i++){
            char a;
            cin >> a;
            if(a%S==0){
                ans += i+1;
            }
        }
    }else{
        int b = 0;
        for(int i=0;i<N;i++){
            char a;
            cin >> a;
            b *= 10;
            int d =a-'0';
            int c = (b + d)%S;
            B[i] = d;
            b = c;
        }
        for(int i=0;i<N;i++){
            A[b] += 1;
            if(b==0){
                b = B[i];
            } else {
                b = (b*10000 - B[i])%S;
            }
        
        }
        A[0] += 1;
        for(int i= 0;i<S;i++){
            ans += A[i]*(A[i]-1)/2;
        }
    }
    
    cout << ans << endl;
    
}