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: Runtime Error)

#include<bits/stdc++.h>
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
 using ll=long long;

int main(){
  int n,p; cin>>n>>p;
  string s; cin>>s;
  if(p==2||p==5){
    ll cnt=0;
    REP(i,n)if((s[i]-'0')%p==0)cnt+=i+1;
    cout<<cnt<<endl;
    return 0;
  }
  assert(false);
  vector<ll> cnt(p);
  ll cur=0,t=1;
  cnt[0]=1;
  REP(i,n){
    cur=t*(s[i]-'0')+cur;
    cur%=p;
    t=10*t%p;
    cnt[cur]++;
  }
  ll ans=0;
  REP(i,p)ans+=cnt[i]*(cnt[i]-1)/2;
  cout<<ans<<endl;
  return 0;
}