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;
#define itn int
#define int unsigned long long
#define mod 1000000007
#define fi first
#define se second
#define be begin()
#define en end()
#define re reverse
#define Set setprecision
#define NO cout<<"NO\n"
#define YES cout<<"YES\n"
#define No cout<<"No\n"
#define Yes cout<<"Yes\n"
const int INFi=2147483647;
const long long INFl=9223372036854775807LL;
signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,p;
    string s;
    int ans=0;
    cin>>n>>p>>s;
    for(int i=0;i<s.size();i++){
	int sum=0;
	for(int j=i;j<s.size();j++){
	    sum*=10;
	    sum+=(s[j]-'0');
	    //cout<<sum<<"\n";
	    if(sum%p==0){
		ans++;
		sum=0;
	    }
	}
    }
    cout<<ans<<"\n";
}