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)

// E - Divisible Substring
#include <bits/stdc++.h>
using namespace std;
#define c2d(c) ((c)-'0') // char to digit
#define modadd(x,y,m) x=(x+(y))%(m)
#define modmul(x,y,m) x=(x*(y))%(m)

int main(){
	int N,P; string S; cin>>N>>P>>S;
	long long total = 0;
	if(10%P == 0){ // P={2,5}
		for(int i=0; i<N; ++i) if(c2d(S[i]) % P == 0) total += i+1;
	}
	else{
		vector<int> cnt(P, 0);
		int r = 0, ten = 1;
		for(int i=N-1; i>=0; --i){
			cnt[r]++;
			modadd(r, c2d(S[i])*ten, P);
			total += cnt[r];
			modmul(ten, 10, P);
		}
	}
	cout << total << endl;
}