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;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
#define rep(i, n) for(ll i = 0;i < n;i++)
#define all(i) i.begin(), i.end()
template<class T, class U> bool cmax(T& a, U b) { if (a<b) {a = b; return true;} else return false; }
template<class T, class U> bool cmin(T& a, U b) { if (a>b) {a = b; return true;} else return false; }

ll modpow(ll a, ll n, ll p) {
  ll res = 1;
  while (n) {
    if (n&1) res = res * a % p;
    a = a * a % p;
    n >>= 1;
  }
  return res;
}

int main() {
  cin.tie(0); ios::sync_with_stdio(false);

  ll n, p;
  string s;
  cin >> n >> p >> s;


  vector<ll> unko(n+1);
  unko[0] = 0;
  ll h = 10%p;
  rep(i, n) {
    ll t = s[i]-'0';
    t %= p;
    unko[i+1] = ((unko[i]*h)%p+t)%p;
  }

  vector<ll> d(n+1);

  rep(i, n+1) {
    d[i] = (unko[i] * modpow(h, max(0LL, i-1), p))%p;
  }

  ll ans = 0;
  vector<ll> cnt(p);
  rep(i, n+1) {
    ans += cnt[d[i]];
    cnt[d[i]]++;
  }
  cout << ans << endl;
}