Problem:
Score : 400 points

Problem Statement

Given is a string S consisting of digits from 1 through 9.

Find the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the following condition:

Condition: In base ten, the i-th through j-th characters of S form an integer that is a multiple of 2019.

Constraints

1 ≤ |S| ≤ 200000

S is a string consisting of digits from 1 through 9.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the condition.

Sample Input 1

1817181712114

Sample Output 1

3

Three pairs - (1,5), (5,9), and (9,13) - satisfy the condition.

Sample Input 2

14282668646

Sample Output 2

2

Sample Input 3

2119

Sample Output 3

0

No pairs satisfy the condition.

Solution in OCaml (4.10.0) (judge verdict: Runtime Error)

open Batteries
(* open Big_int *)
(* 前からst文字を切り抜く *)
let slice st str = String.init (String.length str - st) (fun a -> String.get str (a+st))
(* 文字列をcharのlistにする *)
let explode s = List.init (String.length s) (String.get s)
let s = read_line () |> explode
let sl = List.length s
let dp = Array.init (sl+1) @@ fun a -> ""

let rec loop old lst i =
  match lst with
  | [] -> dp.(i) <- old
  | first :: rest -> dp.(i) <- old; loop (old ^ (Char.escaped first)) rest (i+1)

let _ = loop "" s 0
let t = Big_int.big_int_of_int 2019
let rec loop1 i =
  let rec loop2 j =
    if j > sl then 0 else
    (
      if Big_int.mod_big_int (slice (String.length dp.(i)) dp.(j) |> Big_int.big_int_of_string) t  = Big_int.zero_big_int then 1 + loop2 (j+2) else loop2 (j+1)
    )
  in if i >= sl-3 then 0 else (loop2 (i+1) + loop1 (i+1))

let _ = loop1 0 |> Printf.printf "%d\n"