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: Accepted)

#include <iostream>
#include <string>
#include <unordered_map>

using namespace std;

int main() {
	ios_base::sync_with_stdio(false);
	int n, p;
	cin>>n>>p;
	string s;
	cin>>s;
	long res = 0;
	if (p == 2 || p == 5) {
		for (int i = 0; i < n; i++) {
			if ((s[i] - '0') % p == 0) {
				res += i + 1;
			}
		}
		cout<<res<<endl;
		return 0;
	}
	unordered_map<int, int> mp;
	mp[0] = 1;
	long lastten = 1;
	long last = 0;
	for (int i = n-1; i >= 0; i--) {
		long cur = ((s[i]-'0')*lastten+last) % p;
		last = cur;
		lastten = lastten * 10 % p;
		auto it = mp.find(cur);
		if (it == mp.end()) {
			mp[cur] = 1;
		} else {
			res += it->second;
			it->second++;
		}
	}
	cout<<res<<endl;
	return 0;
}