#include "bits/stdc++.h"
using namespace std;

const int N=2e5+20;

int n,mod,m[N];
long long dp[N],p[N],ans;
char a[N];

int main()
{
	scanf("%d%d %s",&n,&mod,a+1);

	if(mod==2 or mod==5)
	{
		for(int i=1;i<=n;i++)
			if((a[i]-'0')%mod==0) ans+=i;
		return 0 * printf("%lld",ans);
	}

	p[0]=1;
	for(int i=1;i<=n;i++)
	{
		p[i]=(p[i-1]*10)%mod;
		dp[i]=(dp[i-1]*10+a[i]-'0')%mod;
	}

	for(int i=n;i>=1;i--)
	{
		dp[i]=(dp[i]*p[n-i])%mod;
		m[dp[i]]++;
		ans+=m[dp[i]]-1;
	}

	printf("%lld",ans+m[0]);
}
### Prompt

Please formulate 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"
using namespace std;

const int N=2e5+20;

int n,mod,m[N];
long long dp[N],p[N],ans;
char a[N];

int main()
{
	scanf("%d%d %s",&n,&mod,a+1);

	if(mod==2 or mod==5)
	{
		for(int i=1;i<=n;i++)
			if((a[i]-'0')%mod==0) ans+=i;
		return 0 * printf("%lld",ans);
	}

	p[0]=1;
	for(int i=1;i<=n;i++)
	{
		p[i]=(p[i-1]*10)%mod;
		dp[i]=(dp[i-1]*10+a[i]-'0')%mod;
	}

	for(int i=n;i>=1;i--)
	{
		dp[i]=(dp[i]*p[n-i])%mod;
		m[dp[i]]++;
		ans+=m[dp[i]]-1;
	}

	printf("%lld",ans+m[0]);
}
```