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;
#define ll long long
#define ull unsigned long long
int main()
{
    ull n;cin>>n;ull p;cin>>p;
    string s;cin>>s;
    reverse(s.begin(),s.end());
    ull a[n];
    for(int i=0;i<n;i++){
        a[i] = 0;
    }
    ull ans = 0;
    ull prev = 0;
    for(ull i=0;i<n;i++){
        ull here = prev * 10;
        ull temp = s[i] - '0';
        here += temp;
        a[i] = here % p;
        prev = a[i];
    }
    ull cnt[p];
    for(ull i=0;i<p;i++){
        cnt[i] = 0;
    }
    for(ull i=0;i<n;i++) {
        cnt[a[i]] ++;
        if(a[i] == 0) {
            ans += cnt[a[i]];
        }
        else if(p > 10) {
            ans += cnt[a[i]];
            ans --;
        }
    }
    cout<<ans<<endl;
    return 0;
}