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: Accepted)

let rec power ( * ) e m n =
  if n <= 0 then e
  else power ( * ) (if n land 1 = 0 then e else m * e) (m * m) (n lsr 1)

let () = Scanf.scanf "%d %d\n%s" @@ fun n p s ->
  if p = 2 || p = 5 then begin
    Printf.printf "%d\n" @@
    Array.fold_left ( + ) 0 @@
    Array.init n @@ fun i ->
      if 0 < (Char.code s.[i] - Char.code '0') mod p
      then 0
      else i + 1
  end else begin
    let ( +^ ) x y = (x + y) mod p in
    let ( *^ ) x y = (x * y) mod p in
    let acc = Array.make (n + 1) 0 in
    for i = 0 to n - 1 do
      acc.(i + 1) <- acc.(i) +^ power ( *^ ) (Char.code s.[i] - Char.code '0') 10 ((p - 2) * (i + 1))
    done;
    let ms = Array.make p 0 in
    Array.iter (fun m -> ms.(m) <- 1 + ms.(m)) acc;
    Printf.printf "%d\n" @@
    Array.fold_right (fun m -> ( + ) @@ m * (m - 1) / 2) ms 0
  end