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;
inline void prep ()
{
  cin.tie (0);
  cin.sync_with_stdio (0);
};
const double pi = acos(-1.);
//printf(%.12lf\n,mid);
// __builtin_popcount(int) count 1's in binary
// get decimal part of number
/*
double intpart;
double thiss = modf((double)(n*m)/(double)k, &intpart);
*/
// round = (int)(num+.5)
long long mod = 998244353;
 
// 32 mil = ~1 sec 
long long mymod(long long num){
  if (num< 0 && !(num%mod == 0)){
    if (mod == 1){
      return 0;
    }
    return mod-abs(num)%mod;
  }else{
    return num%mod;
  }
}
 
/*string sentence = And I feel fine...;
    istringstream iss(sentence);
*/  
const int biggg = 2000000000;
const long long bigggest = 1000000000000000000LL;
const double eps = .0000000001; 

int main ()
{
  prep();
  int n, p;
  string s;
  cin >> n >> p >> s;
  long long sums[n];
  long long curpow = 1;
  int freqs[p] = {0};
  for (int i=s.size()-1; i>=0; i--){
      sums[i] = curpow*(s[i]-'0');
      curpow *= 10;
      curpow %= p;
  }

  for (int i=0; i<n; i++) {
      if (i > 0) sums[i] += sums[i-1];
      sums[i] %= p;
      freqs[sums[i]]++;
  }
  long long cnt = 0;
  long long cursum = 0;
  for (int i=0; i<s.size(); i++){
      cnt += freqs[cursum];
      freqs[sums[i]]--;
      cursum = sums[i];
  }
  cout << cnt;
  return 0;
}