#include <bits/stdc++.h>
int N, P, x;
long long ans;
char S[200005];
int m[10005];
void test() {
	std::cin >> S + 1;
	long long ans = 0;
	for (int i = 1; i <= N; ++i) {
		int x = S[i] & 15;
		if (x % P == 0) ans += i;
	}
	std::cout << ans << '\n';
}
int main() {
	std::cin >> N >> P;
	if (P == 2 || P == 5) return test(), 0;
	std::cin >> S + 1;
	m[0] = 1;
	int c = 1;
	std::reverse(S + 1, S + N + 1);
	for (int i = 1; i <= N; ++i) {
		x = x + c * (S[i] & 15), x %= P, c = c * 10 % P;
		ans += m[x]; ++m[x];
	}
	std::cout << ans << '\n';
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

Examples

Input

4 3
3543


Output

6


Input

4 2
2020


Output

10


Input

20 11
33883322005544116655


Output

68

### Response

```cpp
#include <bits/stdc++.h>
int N, P, x;
long long ans;
char S[200005];
int m[10005];
void test() {
	std::cin >> S + 1;
	long long ans = 0;
	for (int i = 1; i <= N; ++i) {
		int x = S[i] & 15;
		if (x % P == 0) ans += i;
	}
	std::cout << ans << '\n';
}
int main() {
	std::cin >> N >> P;
	if (P == 2 || P == 5) return test(), 0;
	std::cin >> S + 1;
	m[0] = 1;
	int c = 1;
	std::reverse(S + 1, S + N + 1);
	for (int i = 1; i <= N; ++i) {
		x = x + c * (S[i] & 15), x %= P, c = c * 10 % P;
		ans += m[x]; ++m[x];
	}
	std::cout << ans << '\n';
}
```