#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
string num;
map<int,int> mp;
int main(){
    mp[0]=1;
    ll ans=0,cnt=0,p=1,x;
    cin>>x>>x>>num;
    if(x==2||x==5){
        for(int i=0;i<num.length();i++){
            if((num[i]-'0')%x==0) ans+=(ll)i+1;
        }
    }
    else{
    for(int i=num.length()-1;i>=0;i--){
        cnt=(cnt+(ll)(num[i]-'0')*p)%x;
        ans+=mp[cnt],mp[cnt]++;
        p=p*10%x;
    }
    }
    printf("%lld",ans);
}

### Prompt

In cpp, your task is to solve 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;
typedef long long ll;
string num;
map<int,int> mp;
int main(){
    mp[0]=1;
    ll ans=0,cnt=0,p=1,x;
    cin>>x>>x>>num;
    if(x==2||x==5){
        for(int i=0;i<num.length();i++){
            if((num[i]-'0')%x==0) ans+=(ll)i+1;
        }
    }
    else{
    for(int i=num.length()-1;i>=0;i--){
        cnt=(cnt+(ll)(num[i]-'0')*p)%x;
        ans+=mp[cnt],mp[cnt]++;
        p=p*10%x;
    }
    }
    printf("%lld",ans);
}

```