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 LuaJIT (2.0.4) (judge verdict: Accepted)

local n = io.read("*n", "*l")
local s = io.read()
local sa = {}
for i = 1, n do
  table.insert(sa, {i, s:sub(i, n)})
end
table.sort(sa, function(x, y) return x[2] < y[2] end)
local lcpa = {}
for i = 1, n do
  lcpa[i] = {}
  for j = 1, n do
    lcpa[i][j] = 0
  end
end
local mmi, mma = math.min, math.max
for i = 1, n - 1 do
  local len = mmi(#sa[i][2], #sa[i + 1][2])
  local c = 0
  for j = 1, len do
    if sa[i][2]:sub(j, j) == sa[i + 1][2]:sub(j, j) then
      c = c + 1
    else break end
  end
  local i1, i2 = sa[i][1], sa[i + 1][1]
  lcpa[i1][i2], lcpa[i2][i1] = c, c
end
for i = 1, n - 2 do
  local prv = lcpa[sa[i][1]][sa[i + 1][1]]
  if 0 < prv then
    for j = i + 1, n - 1 do
      local nxt = lcpa[sa[j][1]][sa[j + 1][1]]
      if nxt == 0 then
        break
      else
        prv = mmi(prv, nxt)
        lcpa[sa[i][1]][sa[j + 1][1]], lcpa[sa[j + 1][1]][sa[i][1]] = prv, prv
      end
    end
  end
end
local ret = 0
for i = 1, n - 1 do
  for j = i + 1, n do
    ret = mma(ret, mmi(j - i, lcpa[i][j]))
  end
end
print(ret)