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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
const int maxn=2e5+10;
#define _rep(i,a,b) for(int i=(a);i<=(b);++i)
int n,p,f[maxn];
long long ans;
char s[maxn];

int main()
{
    #ifdef __LOCAL__
        freopen("1.in","r",stdin);
        freopen("1.out","w",stdout);
    #endif
    scanf("%d%d%s",&n,&p,s+1);
    if(p==2||p==5)
    {
        _rep(i,1,n)
            if((s[i]-'0')%p==0)
                ans+=i;
    }
    else
    {
        f[0]=1;
        int k=1,t=0;
        for(int i=n;i>=1;i--)
        {
            t=(t+(s[i]-'0')*k)%p;
            ans+=f[t];
            f[t]++;
            k=k*10%p;
        }
    }    
    cout<<ans<<endl;
    
    return 0;
}