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: Wrong Answer)

Scanf.sscanf (read_line ()) "%d %d" (fun n p ->
    let s = read_line () in
    if p mod 2 = 0 || p mod 5 = 0 then (
        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"
    ) else (
        let dp = Array.make p 0 in
        let rec loop i mul shift acc =
            if i = n then acc else
                let c = int_of_char s.[i] - int_of_char '0' in
                let d = (c * mul + shift) mod p in
                let () = dp.(d) <- dp.(d) + 1 in
                let mul = (mul * 10) mod p in
                let nextshift = (shift * 10 + c) mod p in
                loop (i + 1) mul nextshift (acc + dp.(shift mod p))
        in
        loop 0 1 0 0 |> Printf.printf "%d\n" 
    )
)