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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include <bits/stdc++.h>

using namespace std;

long long solve_1 (int N, int P, string S) {
  long long cnt = 0;

  for (int i = N - 1; i >= 0; i--) {
    if ((S[i] - '0') % P == 0) cnt += i + 1;
  }

  return cnt;
}

long long solve_2 (int N, int P, string S) {
  auto MOD = [&](int a) {
    int res = a % P;
    if (res < 0) res += P;
    return res;
  };

  long long cnt = 0;
  map<int, int> dp;
  dp[0] = 1;
  int x = 0;
  int ten = 1;
  for (int i = N - 1; i >= 0; i--) {
    x = MOD(x + MOD(S[i] - '0') * ten);
    cnt += dp[x];
    dp[x]++;
    (ten *= 10) %= P;
  }

  return cnt;
}

/**
 * main
 */
int main () {
  int N, P; cin >> N >> P;
  string S; cin >> S;

  cout << ((P == 2 || P == 5) ? solve_1(N, P, S) : solve_2(N, P, S))
       << '\n';
}