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 (4.02.3) (judge verdict: Accepted)

let () =
    let main () =
        let n = int_of_string (read_line ()) in
        let s = read_line () in

        let rec loop_l l acc =
            if l = n then acc else (
                let nn = n - l in
                let a = Array.make nn 0 in
                let rec loop_i i j =
                    if i < nn then (
                        let rec loop_j j =
                            if i + j = nn || s.[j + l] != s.[i + j + l] then j else loop_j (j + 1)
                        in
                        let j = loop_j j in
                        a.(i) <- j;
                        if j = 0 then loop_i (i + 1) j else
                            let rec loop_k k =
                                if i + k = nn || k + a.(k) >= j then k else
                                    let () = a.(i + k) <- a.(k) in
                                    loop_k (k + 1)
                            in
                            let k = loop_k 1 in
                            loop_i (i + k) (j - k)
                    )
                in
                loop_i 1 0;
                let rec loop i acc =
                    if i = nn then acc else loop (i + 1) (max acc (min i a.(i)))
                in
                let acc = loop 1 acc in
                loop_l (l + 1) acc
            )
        in
        let r = loop_l 0 0 in
        Printf.printf "%d\n" r
    in
    main ()