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++ (GCC 9.2.1) (judge verdict: Runtime Error)

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

long long N, P;
string S;

long long memo[200005];
map<long long, long long>pcount;
long long kaidan[10000000];

long long modpow(long long ori, long long po){
	long long res = 1;
	while(po > 0){
		if(po&1){
			res *= ori;
			res %= P;
		}
		ori *= ori;
		ori %= P;
		po >>= 1;
	}
	return res;
}


int main(){
	cin >> N >> P >> S;
	if(P == 2 || P == 5){
		return 1;
	}

	for(long long i = 0; i < N; i++) {
		memo[i+1] = memo[i] + (S[N-1-i] - '0') * modpow(10, i);
		memo[i+1] %= P;
		pcount[memo[i+1]]++;
	}
	//for(long long i = 0; i <= N; i++) {
	//	cout << "memo[" << i << "] = " << memo[i] << endl;
	//}


	long long ans = 0;
	for(auto item: pcount) {
		//cout << item.first << ", " << item.second << endl;
		if(item.first == 0) {
			ans += (item.second * (item.second - 1))/2 + item.second;
		} else {
			ans += (item.second * (item.second - 1))/2;
		}
	}
	cout << ans << endl;
}