#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
using namespace std;
const int maxn=1e6+10;
int n,p;
long long ans,dp[10][maxn];
char s[maxn];
map<int,int>num;
int main()
{
	scanf("%d%d%s",&n,&p,s+1);
	if (p==2)
	{
		for (int i=1;i<=n;i++) if ((s[i]-'0')%2==0) ans+=i;
		printf("%lld\n",ans);
	}
	else if (p==5)
	{
		for (int i=1;i<=n;i++) if ((s[i]-'0')%5==0) ans+=i;
		printf("%lld\n",ans);
	}
	else 
	{
		int now=0,mi=1; num[0]=1;
		for (int i=n;i>=1;i--)
		{
			now=(now+(s[i]-'0')*mi%p)%p;
			ans+=num[now]; num[now]++; mi=mi*10%p;
		}
		printf("%lld\n",ans);
	}
return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
using namespace std;
const int maxn=1e6+10;
int n,p;
long long ans,dp[10][maxn];
char s[maxn];
map<int,int>num;
int main()
{
	scanf("%d%d%s",&n,&p,s+1);
	if (p==2)
	{
		for (int i=1;i<=n;i++) if ((s[i]-'0')%2==0) ans+=i;
		printf("%lld\n",ans);
	}
	else if (p==5)
	{
		for (int i=1;i<=n;i++) if ((s[i]-'0')%5==0) ans+=i;
		printf("%lld\n",ans);
	}
	else 
	{
		int now=0,mi=1; num[0]=1;
		for (int i=n;i>=1;i--)
		{
			now=(now+(s[i]-'0')*mi%p)%p;
			ans+=num[now]; num[now]++; mi=mi*10%p;
		}
		printf("%lld\n",ans);
	}
return 0;
}
```