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)

// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#include<chrono>
#include<iomanip>
#include<map>
#include<set>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
void init_io(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << setprecision(10);
}
signed main(){
  init_io();
  ll n,p,tmp=0,ans=0;
  string s;
  cin >> n >> p >> s;
  map<ll,ll> mp;
  mp[0]++;
  vector<ll> sum(n+1,0);
  for(int i=0;i<n;i++){
    tmp += s[i]-'0';
    tmp %= p;
    ans += mp[tmp];
    mp[tmp]++;
    tmp *= 10;
    tmp %= p;
  }
  cout << ans << endl;
}