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)

#line 1 "template.cpp"
#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) (v).begin(), (v).end()
#define coutd(n) cout << fixed << setprecision(n)
#define ll long long int
#define vl vector<ll>
#define vi vector<int>
const int INF = 2147483647;
const ll MOD = 1000000007;

using namespace std;

ll power(ll base, long long int n, ll p) {
  ll res = 1;
  while (n) {
    if (n & 1) {
      res = (res * base) % p;
    }
    base = (base * base) % p;
    n /= 2;
  }
  return res % p;
}

ll inv(ll base, ll p) { return power(base, p - 2, p); }

int main() {
  ll n, p;
  cin >> n >> p;
  string s;
  cin >> s;

  vector<ll> count(p, 0LL);
  ll cur = 0;
  ll inv10 = inv(10LL, p);

  REP(i, n) {
    ll num = s[i] - '0';
    cur = (cur * 10LL + num) % p;
    count[(cur * power(inv10, i, p)) % p]++;
  }

  ll ans = count[0];
  REP(i, p) {
    if (count[i] >= 2) {
      ans += (count[i] * (count[i] - 1LL)) / 2LL;
    }
  }
  cout << ans << endl;
}