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 Fortran (gfortran v4.8.4) (judge verdict: Accepted)

program who_says_a_pun
  implicit none
  integer :: n, i, j, ans = 0, z(5000) = 0
  character(5000) :: s
  read(*,*) n
  read(*,*) s
  do i = 1, n
    z(1:n-i+1) = z_algorithm(s(i:n))
    do j = 2, n-i+1
      ans = max(ans,min(j-1,z(j)))
    end do
  end do
  write(*,'(i0)') ans
contains
  function z_algorithm(s) result(z)
    implicit none
    character(*) :: s
    integer :: z(len_trim(s)), n, i, j, k
    z = 0
    n = len_trim(s)
    z(1) = n
    i = 1
    j = 0
    do while (i < n)
      do while (i+j < n .and. s(j+1:j+1) == s(i+j+1:i+j+1))
        j = j+1
      end do
      z(i+1) = j
      if (j == 0) then
        i = i+1
        cycle
      end if
      k = 1
      do while (i+k < n .and. k+z(k+1) < j)
        z(i+k+1) = z(k+1)
        k = k+1
      end do
      i = i+k
      j = j-k
    end do
  end
end program who_says_a_pun