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"
using namespace std;
typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<ll ,ll> pll;
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define all(X) (X).begin(), (X).end()
#define sll(n) scanf("%lld",&n)
#define sll2(x,y) scanf("%lld%lld",&x,&y)
#define sll3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define REP(i,x,y) for(ll i = x;i <= y;++i)
#define debug(x) cerr<<#x<<"::"<<x<<endl
#define debug2(x,y) cerr<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\n"
#define debug3(x,y,z) cerr<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<"\n"
#define debug4(x,y,z,w) cerr<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<"\t"<<#w<<" :: "<<w<<"\n"
ll power(ll x,ll y,ll z) {
	ll ret = 1;
	while(y > 0) {
		if(y & 1) ret = (ret*x)%z;
		x = (x*x)%z;
		y >>= 1;
	}
	return ret;
}
int main() {
	
	ll n, p;	
	cin >> n >> p;
	string s;
	cin >> s;
	vector<ll>mod(p,0);
	mod[0] = 1;
	ll val = 0, ans = 0, po = 1;
	for(ll i = n;i >= 1;--i) {
		val = (val + (s[i - 1] - '0') * po) % p;
		ans += mod[val];
		po = (po * 10) % p;
		// debug3(i,val,mod[val]);
		mod[val]++;
	}
	if(p == 2 || p == 5) {
		ans = 0;
		REP(i,1,n) {
			if((s[i - 1] - '0') % p == 0) ans += i;
		}
	}
	cout << ans << "\n";

	return 0;
}