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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std; 
#define LIM 1000003

#define M 1000000007
#define inf 99999999999999999LL	//long long inf
typedef long long ll;

#define sc(x) scanf("%d",&x);
#define sc2(x,y) scanf("%d%d",&x,&y);
#define sc3(x,y,z) scanf("%d%d%d",&x,&y,&z);
#define scl(x) scanf("%lld",&x);
#define scl2(x,y) scanf("%lld%lld",&x,&y);
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z);

#define pb push_back
#define mp make_pair
#define fi first
#define se second 
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;}


#define debug(x) cerr<<#x<<" :: "<<x<<"\n";
#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,a) cerr<<#x<<" :: "<<x<<"\t"<<#y<<" :: "<<y<<"\t"<<#z<<" :: "<<z<<"\t"<<#a<<" :: "<<a<<"\n";



int n, p;
string s;
int main() {
	int i, j, t;
	sc2(n, p);
	cin >> s;
	long long int answer = 0;
		
	//debug(s);
	if (p == 2 || p == 5) {
		for(i=0; i<n; i++) {
			int ele = s[i] - '0';
			if(ele % p == 0)
				answer += (i+1);
		}	
	} else {
		int curMod = 0;
		map< int, int >mm;
		mm[curMod]++;
		int powTen = 1LL;
		for(i=0; i<n; i++) {
			int ele = s[i] - '0';
			curMod = (curMod +  ele * powTen)%p;
			powTen = (powTen * 10 )%p;
			answer = (answer + mm[curMod]);
			mm[curMod]++;
		}
	}
	cout << answer << endl;		
	
	return 0;
}