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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
char a[200005];
int n,p,s[200005],cnt[200005];
long long ans=0;
int main(){
	scanf("%d%d%s",&n,&p,a+1);
	if(p==2){
		for(int i=n;i>=1;i--)if(a[i]%2==0)ans+=i;
		cout<<ans;
		return 0;
	}
	if(p==5){
		for(int i=n;i>=1;i--)if(a[i]=='0'||a[i]=='5')ans+=i;
		cout<<ans;
		return 0;
	}
	for(int i=n,pp=1;i>=1;i--,pp=pp*10%p)s[i]=(1ll*(a[i]-'0')*pp%p+s[i+1])%p;
	cnt[0]=1;
	for(int i=n;i>=1;i--){
	//	cout<<s[i]<<' ';
		ans+=cnt[s[i]];
		cnt[s[i]]++;
	}
	cout<<ans;
}