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 OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

Scanf.sscanf (read_line ()) "%d %d" (fun n p ->
    let s = read_line () in
    let dp = Array.make_matrix 2 p 0 in
    let rec loop i acc =
        if i = n then acc else (
            let cur = i mod 2 in
            let next = 1 - cur in
            let c = int_of_char s.[i] - int_of_char '0' in
            for j = 0 to p - 1 do dp.(next).(j) <- 0 done;
            for j = 0 to p - 1 do
                let v = (j * 10 + c) mod p in
                dp.(next).(v) <- dp.(next).(v) + dp.(cur).(j)
            done;
            dp.(next).(c mod p) <- dp.(next).(c mod p) + 1;
            loop (i + 1) (acc + dp.(next).(0))
        )
    in
    loop 0 0 |> Printf.printf "%d\n"
)