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>) {
    val (n, p) = readLine()!!.split(' ').map { it.toInt() }
    val s = readLine()!!.reversed()

    if (10 % p == 0) {
        (0 until n)
                .filter { (s[it] - '0') % p == 0 }
                .map { n - it }
                .sum()
                .let { println(it) }
        return
    }

    val a = IntArray(n)
    var x = 1
    for (i in 0 until n) {
        a[i] = (s[i] - '0') * x % p
        x = x * 10 % p
    }

    val cs = IntArray(n + 1)
    for (i in 0 until n) {
        cs[i + 1] = (cs[i] + a[i]) % p
    }

    var ans = 0L
    val cnt = IntArray(p)
    for (i in 0..n) {
        ans += cnt[cs[i]]++
    }
    println(ans)
}