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>
using namespace std; 
typedef long long ll;
#define pb push_back
#define fi first
#define se second 
int cnt[10004];
int a[200005];
int b[200005];
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n,p;
	cin>>n>>p;
	if(p==2 or p==5){
		ll ans = 0;
		for(ll i=0;i<n;i++){
			char c; 
			cin>>c; 
			int x = c-'0';
			if(x%p==0)ans+=(i+1);
		}
		cout<<ans<<endl;
	}
	else{
		cnt[0] = 1;
		ll ans = 0;
		ll lst = 0;
		for(int i=0;i<n;i++){
			char c; 
			cin>>c; 
			int x = c-'0';
			a[i] = x;
		}
		ll p10 = 1;
		for(int i=n-1;i>=0;i--){
			lst+=a[i]*p10; 
			p10*=10;
			p10%=p;
			lst%=p;
			ans+=cnt[lst];
			cnt[lst]++;
		}
		cout<<ans<<endl;
	}
	return 0;
}