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;
typedef long long ll;
const int N = 2e5+5;
const int maxn = 2e5+5;

vector <int> s[N];
ll f[N];

void f_table(int n){
	f[0] = 0;
	for(ll i = 1;i <= n;++i) f[i] = f[i-1]+i;
	return ;
}

int main() {
	int n,p,x,sum;
	scanf("%d%d",&n,&p);
	f_table(n);
	for(int i = 1;i <= n;++i) scanf("%1d",&x),(sum = sum*10+x)%=p,s[sum].push_back(i);
	
	ll ans = 0;
	if(s[0].size()) ans += f[s[0].size()];
	for(int i = 1;i < p;++i){
		if(s[i].size()) ans += f[s[i].size()-1];
	}
	cout << ans << endl;
	return 0;
}