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<fstream>
#include<string>
#include<vector>
#include<utility>
#include<algorithm>
#include<climits>
#include<set>
#include<map>
#include<cmath>
#include<iomanip>
#include<iterator>
#include<queue>
#include<stack>
#include<cctype>
#include<deque>
#include<time.h>
#include<bitset>
#include<random>
#include<unordered_set>
#include<unordered_map>
#include<sstream>
#include<random>

//by Skeef79

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
#define pb push_back
#define en '\n'
#define for0(i,n) for(int i = 0;i<n;i++)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define vec vector
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;



vector<ll> getdiv(int k) {
	vector<ll> res;
	for (int i = 1; i * i <= k; i++) {
		if (k % i == 0) {
			res.push_back(i);
			res.push_back(k / i);
			if (i == k / i)res.pop_back();
		}
	}
	sort(res.begin(), res.end());
	return res;
}


int main()
{
	

	int n, p;
	string s;
	cin >> n >> p >> s;
	ll ans = 0;
	if (p == 2) {
		for (int i = 0; i < n; i++) {
			if ((s[i] - '0') % 2 == 0)ans += i + 1;
		}
	}
	else if(p==5){
		for (int i = 0; i < n; i++) {
			if ((s[i] - '0') % 5 == 0) ans += i + 1;
		}
	}
	else {
		reverse(s.begin(), s.end());
		vector<ll>v(n), pref(n);
		ll mul = 1;
		for (int i = 0; i < s.length(); i++) {
			int x = s[i] - '0';
			
			x = (x * mul) % p;
			
			v[i] = x;
			mul = (mul * 10) % p;
		}
		pref[0] = v[0];
		for (int i = 1; i < n; i++) {
			pref[i] = v[i] + pref[i - 1];
		}
		int k = 0;

		map<ll, ll> mp;
		mp[0]++;
		for (int i = 0; i < n; i++) {
			ans += mp[pref[i] % p];
			mp[pref[i] % p]++;
		}
		
	}
	cout << ans;
}