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 int ll;
typedef pair<ll, ll> P;
typedef vector<ll> VI;
typedef vector<VI> VVI;
const ll MOD = 1000000007;
const ll INF = 2147483647;
const ll LINF = 9223372036854775807;
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()

ll power(ll x, ll y, ll p) {
	if (y == 0)
		return 1;
	else if (y == 1)
		return x % p;
	else if (y % 2 == 0) {
		ll pow = power(x, y / 2, p);
		return (pow * pow) % p;
	}
	else {
		ll pow = power(x, y / 2, p);
		return (((pow * pow) % p) * x) % p;
	}
}

int main() {
	int n, p; cin >> n >> p;
	string s; cin >> s;
	ll ans = 0;
	if(p==2||p==5){
		REP(i,n){
			if((s[i]-'0')%p==0)
				ans+=i+1;
		}
	}
	else{
		reverse(ALL(s));
		VI r(n+1,0); 
		REP(i,n)r[i+1]=(r[i]+(s[i]-'0')*power(10,i,p))%p;
		VI f(p,0);
		REP(i,n+1)f[r[i]]++;
		REP(i,p)ans+=f[i]*(f[i]-1)/2;
	}
	cout << ans << endl;
	return 0;
}