Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(k,i,n) for(ll i=k;i<n;++i)
#define umi2357 	ios::sync_with_stdio(0);cin.tie(0);
const ll maxn=2e5+7;
ll n,p,ans=0,cnt[maxn],res,dp[maxn],a[maxn];
string s;
int main(void){
    cin>>n>>p;
    cin>>s;
    res=1;
    if(p==2||p==5){
        rep(0,i,n){
            if((s[i]-'0')%p==0)ans+=i+1;
        }
    }else{
        rep(0,i,n){
            ll x=n-1-i;
            a[x]=s[x]-'0';
            dp[x]=(dp[x+1]+a[x]*res)%p;
            res=(res*10)%p;
            ++cnt[dp[x]];
        }
        rep(0,i,n){
            ll x=cnt[i];
            ans+=x*(x-1)/2;
        }
        ans+=cnt[0];
    }
    cout<<ans;
}