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: Accepted)

#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;

int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    int n,p;
    cin >> n >> p;
    string s;
    cin >> s;
    if(p==2){
    	ll ans=0;
    	for(int i=s.length()-1;i>=0;i--){
    		if((s[i]-'0')%2==0)ans+=i+1;
    	}
    	cout << ans << endl;
    	return 0;
    }
    if(p==5){
    	ll ans=0;
    	for(int i=s.length()-1;i>=0;i--){
    		if((s[i]-'0')%5==0)ans+=i+1;
    	}
    	cout << ans << endl;
    	return 0;
    }
    ll a[n],b=(s[n-1]-'0')%p,c=10;a[n-1]=b;
    for(int i=s.length()-2;i>=0;i--){
        (b+=c*(s[i]-'0'))%=p;
        a[i]=b;
        (c*=10)%=p;
    }
    ll num[p];
    for(int i=0;i<p;i++)num[i]=0;
    num[a[0]]=1;
    ll ans=0;
    for(int i=1;i<n;i++){
        ans+=num[a[i]];
        num[a[i]]+=1;
    }
    ans+=num[0];
    cout << ans << endl;
    return 0;
}