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)

fun main(args: Array<String>) = abc158e()

fun abc158e() {
    val (n, p) = readLine()!!.split(' ').map { it.toInt() }
    val s = readLine()!!.toCharArray().map { it.toString().toInt() }.toIntArray()

    var answer = 0L

    when (10 % p) {
        0 -> {
            var cnt = 1
            for (i in s) {
                if (i % p == 0) answer += cnt++
            }
        }
        else -> {
            val mods = IntArray(n + 1)
            var ten = 1
            for ((index, i) in s.withIndex()) {
                mods[index + 1] = (mods[index] + (i * ten % p)) % p
                ten *= 10
                ten %= p
            }

            val left = IntArray(p)
            for (m in mods) answer += left[m]++
        }
    }

    println(answer)
}