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 <iostream>
#include <vector>
#include <algorithm>
#include <random>
#include <map>
using namespace std;
typedef long long int ll;

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n,p; cin >> n >> p;
	string s; cin >> s;
	ll ans=0;
	ll now=0;
	if(p==2){
		reverse(s.begin(),s.end());
		for(int i=0;i<n;i++){
			if((s[i]-'0')%2==0){
				ans+=(n-i);
			}
		}
		cout << ans << endl;
		return 0;
	}else if(p==5){
		reverse(s.begin(),s.end());
		for(int i=0;i<n;i++){
			if((s[i]-'0')%5==0){
				ans+=(n-i);
			}
		}
		cout << ans << endl;
		return 0;
	}
	map<ll,ll> mp;
	mp[0]++;
	for(int i=0;i<n;i++){
		now*=10;
		now+=(s[i]-'0');
		now%=p;
		mp[now]++;
	}
	for(auto p:mp){
		//cout << p.first << "  " << p.second << endl; 
		ans+=(p.second)*(p.second-1)/2;
	}
	cout << ans << endl;
}