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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;

public class Solution
{
    public static void Main()
    {
        var vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
        var N = vals[0];
        var P = vals[1];
        var S = Console.ReadLine().Select(c => c - '0').ToArray();

        var counts = new long[N];
        for (int i = N - 1; i >= 0; i--)
        {
            var n = 0;
            long count = 0;
            for (int j = i; j < N; j++)
            {
                n = n * 10 + S[j];
                n %= P;
                if (n == 0)
                {
                    count++;
                    if (j + 1 < N)
                    {
                        count += counts[j + 1];
                        break;
                    }
                }
            }
            counts[i] = count;
        }

        Console.WriteLine(counts.Sum());
    }
}