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;

string s;
int n, p;
map<int,int> freq;
long long delta = 1;

#define f(i,a,b) for (int i = a; i < b; i++)

long long modpow(long long a, long long b) {
  	long long res = 1;
  	while (b) {
    	if (b&1) res = res*a%p;
      	a = a*a%p;
      	b /= 2;
    }
	return res;
}

int main() {
  	ios_base::sync_with_stdio(0);
  	cin.tie(0);
	cin>>n>>p>>s;
  
  	if (p == 2) {
      	long long ans = 0;
    	f(i,0,n) {
          	if ((s[i]-'0')%2 == 0) {
          		ans += i+1;
        	}
        }
      	cout << ans << endl;
    } else {
      	long long rho = 0, ans = 0;
		f(i,0,n) {
    		rho = (rho*10%p + (s[i]-'0')) % p;
          	long long rhodelta = rho * modpow(delta, p-2) % p;
          	if (!rhodelta) ans++;
          	ans += freq[rhodelta]++;
    		delta = delta * 10 % p;
  	  	}
      	cout << ans << endl;
    }
  	
	return 0;
}