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: Wrong Answer)

import io.StdIn._

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

  val S = readLine.toCharArray.map(_.toInt - 48)

  val d = Array.fill[Int](N + 1)(0)
  var ten = 1
  for (i <- N - 1 to 0 by -1) {
    val a = S(i) * ten % P
    d(i) = (d(i + 1) + a) % 3
    ten = ten * 10 % P
  }

  val cnt = Array.fill[Int](P)(0)
  var ans = 0L
  for (i <- N to 0 by -1) {
    ans = ans + cnt(d(i))
    cnt(d(i)) = cnt(d(i)) + 1
  }

  println(ans)
}