Problem:
Score : 500 points

Problem Statement

N players will participate in a tennis tournament. We will call them Player 1, Player 2, \ldots, Player N.

The tournament is round-robin format, and there will be N(N-1)/2 matches in total.
Is it possible to schedule these matches so that all of the following conditions are satisfied? If the answer is yes, also find the minimum number of days required.

Each player plays at most one matches in a day.

Each player i (1 \leq i \leq N) plays one match against Player A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} in this order.

Constraints

3 \leq N \leq 1000

1 \leq A_{i, j} \leq N

A_{i, j} \neq i

A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} are all different.

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} \ldots A_{1, N-1}
A_{2, 1} A_{2, 2} \ldots A_{2, N-1}
:
A_{N, 1} A_{N, 2} \ldots A_{N, N-1}

Output

If it is possible to schedule all the matches so that all of the conditions are satisfied, print the minimum number of days required; if it is impossible, print -1.

Sample Input 1

3
2 3
1 3
1 2

Sample Output 1

3

All the conditions can be satisfied if the matches are scheduled for three days as follows:

Day 1: Player 1 vs Player 2

Day 2: Player 1 vs Player 3

Day 3: Player 2 vs Player 3

This is the minimum number of days required.

Sample Input 2

4
2 3 4
1 3 4
4 1 2
3 1 2

Sample Output 2

4

All the conditions can be satisfied if the matches are scheduled for four days as follows:

Day 1: Player 1 vs Player 2, Player 3 vs Player 4

Day 2: Player 1 vs Player 3

Day 3: Player 1 vs Player 4, Player 2 vs Player 3

Day 4: Player 2 vs Player 4

This is the minimum number of days required.

Sample Input 3

3
2 3
3 1
1 2

Sample Output 3

-1

Any scheduling of the matches violates some condition.

Solution in LuaJIT (2.0.4) (judge verdict: Time Limit Exceeded)

local n = io.read("*n")
local t = {}
local pos = {}
for i = 1, n do
  pos[i] = 1
  t[i] = {}
  for j = 1, n - 1 do
    t[i][j] = io.read("*n")
  end
end

local ret = 0
local done = {}
while true do
  ret = ret + 1
  local proceed = false
  local remain = false
  for i = 1, n do done[i] = false end
  for i = 1, n do
    if pos[i] < n and not done[i] then
      remain = true
      local nxt = t[i][pos[i]]
      if t[nxt][pos[nxt]] == i and not done[nxt] then
        proceed = true
        pos[i] = pos[i] + 1
        pos[nxt] = pos[nxt] + 1
        done[i] = true
        done[nxt] = true
      end
    end
  end
  if not remain then
    if not proceed then
      ret = ret - 1
    end
    break
  end
  if not proceed then
    ret = -1
    break
  end
end
print(ret)