Problem:
Score : 500 points

Problem Statement

Given is a string S of length N.

Find the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping.

More formally, find the maximum positive integer len such that there exist integers l_1 and l_2 ( 1 \leq l_1, l_2 \leq N - len + 1 ) that satisfy the following:

l_1 + len \leq l_2

S[l_1+i] = S[l_2+i] (i = 0, 1, ..., len - 1)

If there is no such integer len, print 0.

Constraints

2 \leq N \leq 5 \times 10^3

|S| = N

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping. If there is no such non-empty string, print 0 instead.

Sample Input 1

5
ababa

Sample Output 1

2

The strings satisfying the conditions are: a, b, ab, and ba. The maximum length among them is 2, which is the answer.
Note that aba occurs twice in S as contiguous substrings, but there is no pair of integers l_1 and l_2 mentioned in the statement such that l_1 + len \leq l_2.

Sample Input 2

2
xy

Sample Output 2

0

No non-empty string satisfies the conditions.

Sample Input 3

13
strangeorange

Sample Output 3

5

Solution in OCaml (4.10.0) (judge verdict: Wrong Answer)

module IntSet = Set.Make (struct
  type t = int
  let compare = compare
end)

let rec upper_bound l r p =
  if r - l <= 1
  then l
  else let m = (l + r) / 2 in
       if p m 
       then upper_bound m r p
       else upper_bound l m p

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 modulo = 2305843009213693951
let ( +^ ) x y = (x + y) mod modulo
let ( -^ ) x y = x +^ (modulo - y)
let ( *^ ) x y = (x * y) mod modulo
let code c = Char.code c - Char.code 'a'

let hash = Array.fold_left (fun h c -> 26 *^ h +^ code c) 0
exception Exists

let () = Scanf.scanf "%d\n%s" @@ fun n s ->
  Printf.printf "%d\n" @@
  upper_bound 0 (n / 2 + 1) @@ fun l ->
    let h = ref @@ hash @@ Array.init l @@ String.get s in
    let h' = ref @@ hash @@ Array.init l @@ fun i -> s.[i + l] in
    let set = ref @@ IntSet.singleton !h in
    try
      if IntSet.mem !h' !set then raise Exists;
      for i = 0 to n - 2 * l - 1 do
        h := 26 *^ !h +^ code s.[i + l] -^ power (power ( +^ ) 0) (code s.[i]) 26 l;
        h' := 26 *^ !h' +^ code s.[i + 2 * l] -^ power (power ( +^ ) 0) (code s.[i + l]) 26 l;
        set := IntSet.add !h !set;
        if IntSet.mem !h' !set then raise Exists;
      done; false
    with Exists -> true