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>
#include <array>
using namespace std;
using ULL = unsigned long long;
using UL = unsigned;
using LL = long long;
#define rep(i, n) for(UL i = 0; i < (n); i++)

struct Problem {
	void Solve() {
		UL N; cin >> N;
		UL P; cin >> P;
		string S; cin >> S;
		reverse(S.begin(), S.end());
		vector<UL> V(N + 1);
		{
			UL p = 1;
			rep(i, N) {
				V[i + 1] = (S[i] - '0') * p;
				V[i + 1] += V[i];
				V[i + 1] %= P;
				p = p * 10 % P;
			}
		}
		vector<UL> F(P);
		ULL ans = 0;
		rep(i, N + 1) {
			ans += F[V[i]];
			F[V[i]]++;
		}
		cout << ans << endl;
	}

	Problem();
};
int main() {
	unique_ptr<Problem> p(new Problem());
	p->Solve();
	return 0;
}
Problem::Problem() {
	cout << fixed << setprecision(10);
}