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>
#define slowanddetermined ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;

const long long mod=1e9+7;
long long inverseMod(long long x,long long mod);
long long nCr(long long n,long long r,long long mod){
    //when r is lesser and is greater
    long long res=1;
    if(r>n-r)r=n-r;
    long long div=1;
    for(long long  i=0;i<r;i++){
        res=(res*(n-i))%mod;
    }

    for(long long i=1;i<=r;i++){
        div=(div*i)%mod;
    }
    return (res*inverseMod(div,mod))%mod;
}


long long power(long long x,long long n,long long mod){
//(x^n)%mod
long long result=1;

    while(n){
        if(n%2)
            result=(result*x)%mod;
        x=(x*x)%mod;
        n/=2;
    }
    return result;
}

long long inverseMod(long long x,long long mod){

return power(x,mod-2,mod);
}


int main()
{
    slowanddetermined

    int n,p;
    cin>>n>>p;
    string s;
    cin>>s;
    long long ans=0;
    if(p==2){
        for(int i=0;i<n;i++){
            if((s[i]-'0')%2==0)ans+=i+1;
        }
    }
    else if(p==5){
        for(int i=0;i<n;i++){
            if(((s[i]-'0')%5)==0)ans+=i+1;
        }
    }
    else{

        map<long long ,long long >mp;
        mp[0]=1;
       long long a;
        long long last=0;
        long long  cur=0;
        long long val;
        for(int i=0;i<n;i++){
            cur*=10;cur+=s[i]-'0';cur%=p;
            val=power(10,n-i-1,p)*cur;
            val%=p;
            mp[val]++;
        }
        for(auto i:mp){
            ans+=(i.second*(i.second-1))/2;
        }
        //ans+=mp[0];



    }
    cout<<ans;





    return 0;

}