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;
using P=pair<int,int>;
using G=vector<vector<int>>;
using ll=long long;

int main(){
    int n,p;
    cin>>n>>p;
    string s;
    cin>>s;
    if(p==2||p==5){
        ll ans=0;
        for(int i=0;i<n;i++){
            if((s[i]-'0')%p==0LL){
                ans +=i+1;
            }
            
        }
        cout<<ans<<endl;
    }
    else{
        vector<ll>U(n+1,0);
        map<int,int>mods;//各mod pに何個値があるか
        mods[0]=1;
        ll ans=0;

        for(int i=n-1;i>=0;i--){
            //各Uを求める
            ll k=pow(10,n-i);
            U[i]=(U[i+1]+(s[i]-'0')*k)%p;
            //cout<<U[i]<<endl;
            ans+=mods[U[i]];
            mods[U[i]]++;
        }

        cout<<ans<<endl;
    }
}