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

// macro
#define rep(i,n) for(int i=0;i<n;i++)
#define ll long long
#define all(v) v.begin(), v.end()

// code starts
int main()
{
  int n,p;
  string s;
  cin>>n>>p>>s;
  ll ans=0;
  int i;
  if(p==2)
  {
    rep(i,n)if((s[i]-'0')%2==0)ans+=(i+1);
  }
  else if(p==5)
  {
    rep(i,n)if((s[i]-'0')%5==0)ans+=(i+1);  
  }
  else
  {
    vector<int> modp(p,0);
    modp[0]=1;
    int mod10=1;
    int modn=0;
    for(i=n-1;i>=0;i--)
    {
      modn=(modn+mod10*(s[i]-'0'))%p;
      ans+=modp[modn];
      modp[modn]++;
      mod10=(mod10*10)%p;
    }
  }
  cout<<ans<<endl;
}

### 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;

// macro
#define rep(i,n) for(int i=0;i<n;i++)
#define ll long long
#define all(v) v.begin(), v.end()

// code starts
int main()
{
  int n,p;
  string s;
  cin>>n>>p>>s;
  ll ans=0;
  int i;
  if(p==2)
  {
    rep(i,n)if((s[i]-'0')%2==0)ans+=(i+1);
  }
  else if(p==5)
  {
    rep(i,n)if((s[i]-'0')%5==0)ans+=(i+1);  
  }
  else
  {
    vector<int> modp(p,0);
    modp[0]=1;
    int mod10=1;
    int modn=0;
    for(i=n-1;i>=0;i--)
    {
      modn=(modn+mod10*(s[i]-'0'))%p;
      ans+=modp[modn];
      modp[modn]++;
      mod10=(mod10*10)%p;
    }
  }
  cout<<ans<<endl;
}

```