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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder
{
    class E
    {
        public static void Main()
        {
            var np = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
            int n = np[0];
            int p = np[1];

            string s = Console.ReadLine();

            long count = 0;

            if (p == 2 || p == 5)
            {
                for (int i = 0; i < s.Length; i++)
                    if ((s[i] - '0') % p == 0)
                        count += i + 1;
            }
            else
            {
                var countMod = new long[p];
                int digit = 1;
                int r = 0;
                for (int i = s.Length - 1; i >= 0; i--)
                {
                    r = (digit * (s[i] - '0') + r) % p;
                    countMod[r]++;
                    digit = digit * 10 % p;
                }

                count = countMod[0];
                for (int i = 0; i < p; i++)
                    count += countMod[i] * (countMod[i] - 1) / 2;
            }
            Console.WriteLine(count);
        }
    }
}