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 <iostream>
#include <map>
using namespace std;
long long pow(long long i, long long P);

int main(void){
  long N,P;
  string _S;
  cin>>N>>P;
  cin>>_S;


  if(P==2 or P==5){
    long long retval=0;
    for(long long i=0;i<N;i++){
      if((_S[i]-'0')%P==0){
        retval+=(i+1);
      }
    }
    cout<<retval<<endl;
    return 0;
  }



  map<long long,long long>d;
  for(long long i=0;i<P;i++){
    d[i]=0;
  }
  d[0]=1;
  long long S=0;
  for(long long i=0;i<N;i++){
    S+=pow(i+1,P)*(_S[N-i-1]-'0');
    d[S%P]=d[S%P]+1;
  }
  long long retval=0;
  for(long long i=0;i<P;i++){
    if(d[i]>0){
      retval+=(d[i]-1)*d[i]/2;
    }
  }
  cout<<retval<<endl;
  return 0;
  return 0;
}

long long pow(long long i,long long P){
  if(i==0) return 1;
  else if(i%2==0){
    return (pow(i/2,P)*pow(i/2,P))%P;
  }
  else{
    return (10*pow(i-1,P))%P;
  }
}