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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

object Problem {
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val P = sc.next().toInt()
        val S = sc.next().map { it - '0' }.toTypedArray()

        var ans = 0L
        if (10 % P == 0) {
            for (i in S.indices) {
                if (S[i] % P == 0) {
                    ans += i + 1
                }
            }
        } else {
            var d = 1
            val R = S.indices.map { i ->
                val r = S[i] * d % P
                d *= 10
                d %= P
                r
            }.reversed()
            val R2 = mutableListOf(0)
            var i = 0
            R.forEach {
                i += it
                i %= P
                R2.add(i)
            }
            for (j in 0 until P) {
                val n = R2.count { it == j }
                ans += (n * (n - 1) / 2)
            }
        }
        pw.println(ans)
    }
}