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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main{
    static MyReader in = new MyReader();

    public static void main(String[] args){
        int N, P; {int[] a = in.ii(); N = a[0]; P = a[1];}
        String S = in.s();
        long ans = 0;
        
        if(P != 2 && P != 5){
            int[] mod = new int[N+1];
            int ten = 1;
            
            for(int i = N-1; i >= 0; i--){
                int n = S.charAt(i) - '0';
                mod[i] = (mod[i+1] + ten * n) % P;
                ten = ten * 10 % P;
            }
            
            int[] count = new int[P];
            for(int i = 0; i <= N; i++){
                count[mod[i]]++;
            }
            
            for(int i = 0; i < count.length; i++){
                ans += (long)(count[i] - 1) * count[i] / 2;
            }        
        } else {
            int tmp = 0;
            for(int i = N-1; i >= 0; i--){
                int n = S.charAt(i) - '0';
                tmp += n % P == 0 ? 1 : 0; 
                ans += tmp;
            }
        }
        
        out.println(ans);
    }

    static class MyReader extends BufferedReader{
        MyReader(){
            super(new InputStreamReader(System.in));
        }

        String s(){
            try{return readLine();}catch(IOException e){return "";}
        }

        String[] ss(){
            return s().split(" ");
        }

        int i(){
            return Integer.parseInt(s());
        }

        int[] ii(){
            String[] ss = ss();
            int[] ii = new int[ss.length];
            for(int j = 0; j < ss.length; j++) ii[j] = Integer.parseInt(ss[j]);
            return ii;
        }

        ArrayList<Integer> li(){
            String[] ss = ss();
            ArrayList<Integer> li = new ArrayList<>();
            for(int j = 0; j < ss.length; j++) li.add(Integer.parseInt(ss[j]));
            return li;
        }

        long l(){
            return Long.parseLong(s());
        }

        long[] ll(){
            String[] ss = ss();
            long[] ll = new long[ss.length];
            for(int j = 0; j < ss.length; j++) ll[j] = Long.parseLong(ss[j]);
            return ll;
        }
    }
}