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)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
typedef long long ll;
const int INF (1<<30);

int mod = 1e9+7;
struct mint {
  ll x;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};

int main() {
  int N, P;
  cin >> N >> P;
  string S;
  cin >> S;
  reverse(all(S));

  auto corner = [&](int a) {
    ll ansC = 0LL;
    rep(i,S.length()) {
      int si = S[i] - '0';
      if (si%a == 0) ansC += (N-i);
    }
    return ansC;
  };


  ll ans = 0LL;
  if (P == 2) {
    ans = corner(2);
  }
  else if (P == 5) {
    ans = corner(5);
  }
  else {
    mod = P;
    vector<mint> sum(N);
    sum[0] = (S[0]-'0') % P;
    for (int i = 1; i < N; i++) {
      mint z = 10;
      sum[i] = (mint)(S[i]-'0')*z.pow(i) + sum[i-1];
    }
    vector<int> tmp(P, 0);
    rep(i, N) {
      tmp[sum[i].x]++;
    }

    rep(i, P) {
      ll y = tmp[i];
      if (i == 0) ans += (y+1) * y / 2;
      else ans += y * (y-1) / 2;
    }
  }
   

  cout << ans << endl;

  return 0;
}