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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
#define all(v) v.begin(),v.end()

using namespace std;
typedef long long ll;




int main()
{
    int N;
    cin >> N;
    string S;
    cin >> S;
    int ans = 0;
    
    for (int i = 1; i < N; i++) {
        for (int j = 0; j < N-i; j++) {
            
            if(S[i] == S[j]){
                int a = 1;
                
                while (j+a<=i && i <= N-a+1) {
                    if(S[i+a-1] == S[j+a-1]) a++;
                    else break;
                }
                //i += a-1;
                j += a-1;
                //cout << a << endl;
                /*
                for (int h = i; h < i+a; h++) {
                    cout << S[h] << " " << h << " ";
                }
                cout << endl;
                for (int h = j; h < j+a; h++) {
                    cout << S[h] << " " << h << " ";
                }
                cout << endl;
                cout << endl;
                 */
                ans = max(a-1, ans);
            }
            
            
            
        }
    }
    
    cout << ans << endl;
    
    
    return 0;
}