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>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n,p;
string s;
int dp[10001];
int val[200001];
int val2[200001];
int to[10001][10];
int rev[10001][10];
ll dp2[2][101];

void solve(){
	int prv=0,nxt=1;
	ll ans=0;
	for(int i=0;i<n;i++){
		dp2[prv][0]++;
		for(int j=0;j<p;j++){
			dp2[nxt][(j*10+(s[i]-'0'))%p]+=dp2[prv][j];
		}
		ans+=dp2[nxt][0];
		swap(prv,nxt);
		memset(dp2[nxt],0,sizeof(dp2[nxt]));
	}
	printf("%lld\n",ans);
	return;
}


int main(void){
	scanf("%d%d",&n,&p);
	
	for(int i=0;i<p;i++){
		for(int j=0;j<min(10,p);j++){
			to[i][j]=(i*10+j)%p;
			rev[to[i][j]][j]=i;
		}
	}
	cin >> s;
	if(p<100){
		solve();
		return 0;
	}
	ll ans=0;
	int vt=1;
	int sum=0;
	for(int i=n-1;i>=0;i--){
		sum+=vt*(s[i]-'0');
		sum%=p;
		dp[sum]++;
		val[i]=sum;
		vt=vt*10%p;
	}
	int cur=0;
	for(int i=n-1;i>=0;i--){
		//printf("%d\n",cur);
		ans+=dp[cur];
		dp[val[i]]--;
		cur=rev[cur][(s[i]-'0')];
	}
	printf("%lld\n",ans);
	return 0;
}