#include <bits/stdc++.h>
using namespace std;

int main() {
  int N, P;
  string S;
  cin >> N >> P >> S;
  reverse(S.begin(), S.end());
  
  if (P == 2 || P == 5) {
    long ans = 0;
    for (int i = 0; i < N; i++) {
      if ((S.at(i) - '0') % P == 0) ans += N - i;
    }
    return cout << ans << "\n", 0;
  }
  
  int dig = 1, sum = 0;
  map<int, int> MA;
  MA[0]++;
  for (int i = 0; i < N; i++) {
    (sum += (S.at(i) - '0') * dig) %= P;
    (dig *= 10) %= P;
    MA[sum]++;
  }
  long ans = 0;
  for (auto ma : MA) ans += (long) ma.second * (ma.second - 1) / 2;
  cout << ans << "\n";
}
### Prompt

Please formulate a Cpp solution to the following problem:

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.

Examples

Input

4 3
3543


Output

6


Input

4 2
2020


Output

10


Input

20 11
33883322005544116655


Output

68

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
  int N, P;
  string S;
  cin >> N >> P >> S;
  reverse(S.begin(), S.end());
  
  if (P == 2 || P == 5) {
    long ans = 0;
    for (int i = 0; i < N; i++) {
      if ((S.at(i) - '0') % P == 0) ans += N - i;
    }
    return cout << ans << "\n", 0;
  }
  
  int dig = 1, sum = 0;
  map<int, int> MA;
  MA[0]++;
  for (int i = 0; i < N; i++) {
    (sum += (S.at(i) - '0') * dig) %= P;
    (dig *= 10) %= P;
    MA[sum]++;
  }
  long ans = 0;
  for (auto ma : MA) ans += (long) ma.second * (ma.second - 1) / 2;
  cout << ans << "\n";
}
```