#include <bits/stdc++.h>
using namespace std;
int main() {
	int n, p; cin >> n >> p;
	string s; cin >> s;
	long long ans = 0;
	if (p == 2 or p == 5) {
		for (int i = 0; i < n; i++) 
			if ((s[i] - '0') % p == 0) ans += i + 1;
	} else {
		map<int, int> mp;
		mp[0] = 1;
		int sum = 0, power = 1;
		for (int i = n - 1; i >= 0; i--) {
			sum = (sum + (s[i] - '0') * power) % p;
			ans += mp[sum % p]++;
			power = power * 10 % p;
		}
	}
	cout << ans;	
}
### Prompt

Create a solution in CPP for 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>
using namespace std;
int main() {
	int n, p; cin >> n >> p;
	string s; cin >> s;
	long long ans = 0;
	if (p == 2 or p == 5) {
		for (int i = 0; i < n; i++) 
			if ((s[i] - '0') % p == 0) ans += i + 1;
	} else {
		map<int, int> mp;
		mp[0] = 1;
		int sum = 0, power = 1;
		for (int i = n - 1; i >= 0; i--) {
			sum = (sum + (s[i] - '0') * power) % p;
			ans += mp[sum % p]++;
			power = power * 10 % p;
		}
	}
	cout << ans;	
}
```