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;
int main(){
  int64_t N, P;
  cin >> N >> P;
  string S;
  cin >> S;
  if(P != 2 && P != 5){
  int64_t dp[N+1];
  for(int i=0; i<N+1; i++){
    dp[i] = 0;
  }
  for(int i=0; i<N; i++){
    int x = S.at(i) - '0';
    if(i == 0){
      dp[i] = x%P;
    }
    else{
      dp[i] = (dp[i-1]*10 + x)%P;
    }
    //cout << dp[i] << endl;
  }
  
  map<int64_t,int64_t> mp;
  for(int i=0; i<N+1; i++){
    if(mp.count(dp[i])){
      mp[dp[i]]++;
    }
    else{
      mp[dp[i]] = 1;
    }
  }
  
  int64_t ans = 0;
  for(auto itr = mp.begin(); itr != mp.end(); itr++){
    ans += (itr->second)*(itr->second-1)/2;
  } 
  cout << ans << endl;
  }
  else{
    int64_t cnt = 0;
    for(int i=0; i<S.size(); i++){
      int x = S.at(i) - '0';
      if(x %P == 0){
        cnt += i+1;
      }
    }
    cout << cnt << endl;
  }
}