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>
#define mod 998244353
#define db(a) cout << a << endl
#define db2(a,b) cout << a << " " << b << endl
#define dbp(a) cout << a.first << " " << a.second << endl
#define adb(a) for(auto i:a) cout << i << " "; cout << endl
#define adb2(a) for(auto i:a) cout << i.first << " " << i.second << endl;
#define fastIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define N 100010
using namespace std;

int main()
{
    fastIO;
    int n, p;
    cin >> n >> p;

    string s;
    cin >> s;

    long long ans = 0, i;
    
    if(p == 2 || p == 5)
    {
        for(i=0;i<n;i++) 
        {
            if((s[i]-'0')%p == 0) ans += (i+1);
        }
    }
    else
    {
        long long rem[p] = {0}, currem = 0, pow10 = 1;
        rem[0] = 1;
        
        for(i=n-1;i>=0;i--)
        {
            currem = (currem + (s[i]-'0')*pow10)%p;
            ans += rem[currem];
            rem[currem]++;
            pow10 = (pow10*10)%p;
        }
    }

    cout << ans << endl;
}