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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>
using namespace std;
typedef long long LL;

LL cnt[20005] = {0},P,now = 0,ans = 0,n;
string s;

LL kpw(LL x,LL y){
	LL ans = 1,tmp = x % P;
	while(y){
		if(y & 1) ans = (ans * tmp) % P;
		tmp = (tmp * tmp) % P;
		y >>= 1;
	}
	return ans;
}

int main(){
	ios::sync_with_stdio(false);
	cin >> n >> P;
	cin >> s; s = ' ' + s;
	
	if(P == 2){
		for(LL i = 1;i <= n;i ++){
			if((s[i] - '0') % 2 == 0) ans += i;
		}
		cout << ans << endl;
		return 0;
	}
	
	if(P == 5){
		for(LL i = 1;i <= n;i ++){
			if((s[i] - '0') % 5 == 0) ans += i;
		}
		cout << ans << endl;
		return 0;
	}
	
	cnt[0] ++;
	for(LL i = n;i >= 1;i --){
		now = (now + ((LL)(s[i] - '0')) * kpw(10,(n - i))) % P;
		ans += cnt[now]; cnt[now] ++;
	}
	cout << ans << endl;
	return 0;
}