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>

#define FAST_IO                       \
	ios_base::sync_with_stdio(false); \
	cin.tie(0);                       \
	cout.tie(0);

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;

const int MOD = 1e9 + 7;

int main()
{

	FAST_IO;

	ll n, p;
	string s;
	cin >> n >> p >> s;
	vector<ll> prefix(n + 1);
	ll cnt = 0;
	for (int i = 1; i <= n; ++i) {
		prefix[i] = prefix[i - 1] * 10 + (s[i - 1] - '0');
		prefix[i] %= p;
	}
	sort(prefix.begin(), prefix.end());
	ll ans = 0;
	for (int i = 1; i <= n;) {
		auto low = lower_bound(prefix.begin(), prefix.end(), prefix[i]);
		auto high = upper_bound(prefix.begin(), prefix.end(), prefix[i]);
		ll k = high - low;
		i += k;
		ans += k * (k - 1) / 2;
	}
	cout << ans;
	return 0;
}