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)

import java.util.Scanner


fun main(args: Array<String>) {
	val scn = Scanner(System.`in`)
	val n = scn.nextInt()
	val p = scn.nextInt()
	val s = scn.next().toCharArray().map{Integer.parseInt(it.toString())}
	var ans = 0L
	if (p==2 || p==5){
		s.forEachIndexed {i, d -> if (d%p==0) ans += i+1 }
	}else {
		val map = mutableMapOf<Int, Long>()
		var crt = s.last()%p
		var k = 1
		map[crt] = 1
		for (i in s.lastIndex-1 downTo 0){
			k = (k*10)%p
			crt = (k*s[i]+crt)%p
			map[crt] = map.getOrElse(crt){0} + 1
		}
		map.values.forEach { ans += it*(it-1)/2 }
		if (map.containsKey(0)){
			ans += map[0]!!
		}
	}
	print(ans)
}