#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
int n,m;
long long tot,ans,k=1;
char s[200010];
map<int,int> sum;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s);
	if(m!=2&&m!=5)
	{
		for(int i=n-1;i>=0;i--)
		{
			ans=(ans+1ll*(s[i]-'0')*k)%m;
			k=(k*10)%m;
			sum[ans]++;
		}
		tot=sum[0];
		for(int i=0;i<=m-1;i++)
		{
			tot+=(1ll*(sum[i]-1)*sum[i])/2;
		} 
	}else{
		for(int i=n-1;i>=0;i--)
		{
			if((s[i]-'0')%m==0)
			{
				tot+=i+1;
			}
		}
	}
	printf("%lld\n",tot);
	return 0;
} 
### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#define mod 1000000007
using namespace std;
int n,m;
long long tot,ans,k=1;
char s[200010];
map<int,int> sum;
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s);
	if(m!=2&&m!=5)
	{
		for(int i=n-1;i>=0;i--)
		{
			ans=(ans+1ll*(s[i]-'0')*k)%m;
			k=(k*10)%m;
			sum[ans]++;
		}
		tot=sum[0];
		for(int i=0;i<=m-1;i++)
		{
			tot+=(1ll*(sum[i]-1)*sum[i])/2;
		} 
	}else{
		for(int i=n-1;i>=0;i--)
		{
			if((s[i]-'0')%m==0)
			{
				tot+=i+1;
			}
		}
	}
	printf("%lld\n",tot);
	return 0;
} 
```