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: Time Limit Exceeded)

#pragma GCC optimize("O2,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
#define pb(X) push_back(X)
#define x first
#define y second
#define all(x) (x).begin(), (x).end()
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
using namespace std;

typedef pair<int, int> pii;

const int INF = 1e9 + 7;
int q;

int main() {
	INIT
	int n, p;
	long long ans = 0;
	string s;
	cin >> n >> p >> s;
	for (int i = 0; i < n; ++i) {
		int cnt = 0;
		for (int j = i; j < n; ++j) {
			cnt = (cnt * 10 + (s[j] - '0')) % p;
			if (cnt == 0) ++ans;
		}
	}
	cout << ans;
	return 0;
}