#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
	int n,p,t=0;
	string s;
	cin>>n>>p;
	cin>>s;
	ll ans=0;
	if(p==5)
	{
		for(int i=0;i<n;i++)
		{
			if((s[i]-'0')%p==0)ans+=(i+1);
		}
	}
	else if(p==2)
	{
		for(int i=0;i<n;i++)
		{
			if((s[i]-'0')%p==0)ans+=(i+1);
		}
	}
	else
	{	map<int, int> mp;
		mp[0]++;int p2=1;
		for(int i=n-1;i>=0;i--)
		{
			t=(t+(s[i]-'0')*p2)%p;
			ans=ans+mp[t];
			mp[t]++;
			p2=(p2*10)%p;
		}
	}
	cout<<ans;
}
### Prompt

Develop a solution in CPP 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>
using namespace std;
typedef long long ll;
int main()
{
	int n,p,t=0;
	string s;
	cin>>n>>p;
	cin>>s;
	ll ans=0;
	if(p==5)
	{
		for(int i=0;i<n;i++)
		{
			if((s[i]-'0')%p==0)ans+=(i+1);
		}
	}
	else if(p==2)
	{
		for(int i=0;i<n;i++)
		{
			if((s[i]-'0')%p==0)ans+=(i+1);
		}
	}
	else
	{	map<int, int> mp;
		mp[0]++;int p2=1;
		for(int i=n-1;i>=0;i--)
		{
			t=(t+(s[i]-'0')*p2)%p;
			ans=ans+mp[t];
			mp[t]++;
			p2=(p2*10)%p;
		}
	}
	cout<<ans;
}
```