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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*
class RollingHash(val N:Int,val S:String){
  val base = 1009L
  val mod = 10000000000009L
  val powBase = Array(N){1L}
  init{
    for(i in 1 until N){
      powBase[i] = powBase[i-1]*base%mod
    }
  }
  fun check(len:Int):Boolean{
    val T = TreeMap<Long,Int>()
    var hash = 0L
    for(i in 0 until len){
      hash = (hash + S[i].toLong()*powBase[len-i-1])%mod
    }
    T.put(hash,0)
    for(i in 0 until N-len){
      hash = (hash - S[i].toLong()*powBase[len-1])*base+S[i+len].toLong()
      hash = hash%mod
      hash = if(hash>=0)hash else hash+mod
      if(T.containsKey(hash)&&i-T[hash]!!>=len){
        return true
      }
      T.put(hash,i+1)
    }
    return false
 }
  fun maxlen():Int{
    var left = 0
    var right = (N/2+1).toInt()
    while(right-left>1){
      val len = (right+left)/2.toInt()
      if(check(len))left = len else right = len
    }
    return left
  }
}
fun main(args:Array<String>){
  val rh = RollingHash(readLine()!!.toInt(),readLine()!!)
  println(rh.maxlen())
}