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 Scala (2.11.7) (judge verdict: Runtime Error)

import io.StdIn._

object Main extends App {
  val line1 = readLine.split(" ").map(_.toInt)
  val N = line1(0)
  val P = line1(1)

  val S = readLine.stripMargin

  val matrix = Array.fill[Array[Int]](N)(Array.fill[Int](N)(0))
  var divisibleCount = 0L

  for (i <- 0 until N) {
    for (j <- (N - i - 1) until N) {

      val remaining = if (j == (N - i - 1)) {
        val singleDigit = S.substring(j, j + 1).toInt
        singleDigit % P
      } else {
        val remainingOnLeft = matrix(i)(j - 1)
        val remainingOnTop = matrix(N - j - 1)(j)
        (remainingOnLeft * 10 + remainingOnTop) % P
      }

      if (remaining == 0) {
        divisibleCount += 1
      }

      matrix(i)(j) = remaining
    }
  }

  println(divisibleCount)
}