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

int cnt[10010];

int32_t main(){
    int n, p; cin >> n >> p;
    string s; cin >> s;
    if(p == 2 || p == 5){
        int ans=0;
        for(int i=0; i<n; ++i) if((s[i]-'0')%p == 0) ans+= i+1;
        cout << ans << endl;
        exit(0); 
    }
    int ans = 0;
    int pre = 0; cnt[0]=1;
    int pow10=1;
    for(int i=n-1; i>=0; --i){
        pre=(pre+(s[i]-'0')*pow10)%p, pow10=(pow10*10)%p;
        ans+=cnt[pre];
        ++cnt[pre];
    }
    cout << ans << endl;
}

### 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>
using namespace std;
#define int long long

int cnt[10010];

int32_t main(){
    int n, p; cin >> n >> p;
    string s; cin >> s;
    if(p == 2 || p == 5){
        int ans=0;
        for(int i=0; i<n; ++i) if((s[i]-'0')%p == 0) ans+= i+1;
        cout << ans << endl;
        exit(0); 
    }
    int ans = 0;
    int pre = 0; cnt[0]=1;
    int pow10=1;
    for(int i=n-1; i>=0; --i){
        pre=(pre+(s[i]-'0')*pow10)%p, pow10=(pow10*10)%p;
        ans+=cnt[pre];
        ++cnt[pre];
    }
    cout << ans << endl;
}

```