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: Accepted)

fun main(args: Array<String>) {
    val (N, P) = listOfInt()
    val S = next().map { it - '0' }
    var ans = 0L
    if (10 % P == 0) { //P = 2, 5
        repeat(N) { r ->
            if (S[r] % P == 0) ans += r + 1
        }
        println(ans)
        return
    }
    val d = IntArray(N + 1)
    var ten = 1
    for (n in (0 until N).reversed()) {
        val a = S[n] * ten % P
        d[n] = (d[n + 1] + a) % P
        ten *= 10; ten %= P
    }
    val c = IntArray(P)
    for (n in (0 .. N).reversed()) {
        ans += c[d[n]]
        c[d[n]] += 1
    }
    println(ans)
}
fun next() = readLine()!!
fun listOfString() = next().split(" ")
fun listOfInt() = listOfString().map(String::toInt)