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 Rust (1.42.0) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

fn z_algorithm(s: &str) -> Vec<usize> {
    let mut res = vec![0; s.len()];
    res[0] = s.len();

    let mut i = 1;

    while i < s.len() {
        let mut p = 0;
        while i + p < s.len() {
            if &s[p..p + 1] != &s[i + p..i + p + 1] {
                break;
            }
            p += 1;
        }
        res[i] = p;
        for j in i + 1..i + p {
            if res[j - i] + j > i + p {
                i = j - 1;
                break;
            }
            res[j] = res[j - i];
        }
        i += 1;
    }

    res
}

fn main() {
    input! {
        n:usize,
        s:String,
    }
    let mut ans = 0_usize;
    for i in 0..n - 1 {
        ans = std::cmp::max(
            *z_algorithm(&s[i..])
                .iter()
                .filter(|&&x| (x <= (s.len() - i) / 2))
                .max()
                .unwrap(),
            ans,
        );
    }
    println!("{}", ans);
}