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;
#define rep(i, n) for(int i=0;i<n;i++)
#define REP(i, n) for(int i=1;i<=n;i++)
typedef long long ll;

int main() {
	ll n, p;
	cin >> n >> p;
	string s;
	cin >> s;
	ll ans = 0;
	if (10 % p == 0) {
		rep(i, n) {
			if ((s[i] - '0') % p == 0) ans += i + 1;
		}
		cout << ans << endl;
		return 0;
	}
	ll ten = 1;
	reverse(s.begin(), s.end());
	vector<ll> m(n + 1, 0);
	REP(i, n) {
		m[i] = m[i - 1] + ten * (s[i - 1] - '0');
		m[i] %= p;
		ten *= 10;
		ten %= p;
	}
	vector<int> t(p, 0);
	rep(i, n + 1) {
		t[m[i]]++;
	}
	rep(i, p) {
		ans += t[i] * (t[i] - 1) / 2;
	}
	cout << ans << endl;
	return 0;
}