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)
#define rrep(i,n) for(int i=1;i<(n);++i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define maxs(a, b) a = max(a, b)
#define mins(a, b) a = min(a, b)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const ll linf = (1ll << 61);
const int inf = 1001001001;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n, p;
	cin >> n >> p;
	string s;
	cin >> s;
	vector<int> sum(n + 1);
	int ten = 1;
	for (int i = n - 1; i >= 0; --i) {
		int a = (s[i] - '0') * ten % p;
		sum[i] = (sum[i + 1] + a) % p;
		ten *= 10; ten %= p;
	}
	if (p == 2 || p == 5) {
		ll ans = 0;
		rep(i, n)
			if ((s[i] - '0') % p == 0) ans += (i + 1);
		printf("%lld\n", ans);
	}
	else {
		vector<int> res(p);
		ll ans = 0;
		for (int i = n; i >= 0; --i) {
			ans += res[sum[i]];
			res[sum[i]]++;
		}
		printf("%lld\n", ans);
	}
	return 0;
}