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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;

int main(){
	int n, p;
  	cin >> n >> p;
  	string s;
  	cin >> s;
  
  	if(10 % p == 0){
    	long long ans = 0;
      	rep(i, n){
        	if((s[i] - '0') % p == 0){
            	ans += i + 1;
            }
        }
      	cout << ans << endl;
       	return 0;
    }
  
  	vector<int> d(n + 1);
  	int ten = 1;
  	for(int i = n - 1; i >= 0;i--){
    	int a = (s[i] - '0') * ten % p;
      	d[i] = (d[i + 1] + a) % p;
      	ten *= 10; ten %= p;
    }
  
  	vector<int> cnt(p);
  	long long ans = 0;
  	for(int i = n; i >= 0; i--){
    	ans += cnt[d[i]];
      	cnt[d[i]]++;
    }
  	cout << ans << endl;
}