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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <deque>
#include <climits>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
//#include <tuple>

#define INF INT_MAX>>1 
#define SIZE 100010
#define MOD 1000000007
#define rep(i, n) for(int i=0; i<(int)(n); i++)
#define rep2(i, from, to) for(int i=(from); i<(int)(to); i++)
#define all(obj) (obj).begin(), (obj).end()
#define lpair pair<ll, ll>
#define vint vector<ll>
#define vinput(name) for(int i=0; i<(int)(name.size()); i++) cin >> name[i];
#define voutput(name) for(int i=0; i<(int)(name.size()); i++) cout << name[i] << endl;
typedef long long ll;

using namespace std;

ll ans = 0;
int main() {
	int n, p;
	cin >> n >> p;
	string s;
	cin >> s;
	if(p == 2 || p == 5) {
		rep(r, n) {
			if((s[r]-'0') % p == 0) {
				ans += r+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);
	ll ans = 0;
	for(int i = n; i >= 0; --i) {
		ans += cnt[d[i]];
		cnt[d[i]]++;
	}
	cout << ans << endl;
	return 0;
}