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 _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)

typedef long long ll;
typedef pair<ll, ll> P;
const ll INF = 1LL<<60;

ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

ll modpow(ll a, ll n, ll mod) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}


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

  int N, P;
  cin >> N >> P;

  string s;
  cin >> s;

  reverse(s.begin(), s.end());

  vector<ll> cnt(P);

  rep(i, P) cnt[i] = 0;
  cnt[0]++;
  ll now_mod = 0;
  ll pow10modP = 1;

  vector<ll> cnt_rec(N);

  rep(i, s.length()){

    ll modP = (now_mod + pow10modP * int(s[i] - '0')) % P;

    cnt[modP]++;
    cnt_rec[i] = modP;

    pow10modP = (10 * pow10modP) % P;

    now_mod = modP;

  }

  // ll inv10P = modinv(10, P);

  // // ll T = int(s[0]-'0') % P;

  // ll T = 0;

  // ll pwP = 1;
  // ll ans = 0;
  // rep(i, s.length()){
    
  //   // cout << T << " " << cnt[T] << endl;
  //   ans += cnt[T];

  //   cnt[cnt_rec[i]]--;
    
  //   // T = ((T + 10 * P - int(s[i]-'0')) % P * inv10P) % P;

  //   pwP = (pwP * 10) % P;
  //   T = (T + pwP * int(s[i]-'0')) % P;

  // }

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

  if(P == 5 || P == 2){
    
    ans = 0;

    rep(i, s.length()){

      int num = s[i]-'0';

      if(num % P == 0){
        ans += s.length()-i;
      }
    }
  }

  cout << ans << endl;  

  return 0;
}