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: Accepted)

local n = io.read("*n")
local edge = {}
local srccnt = {}
for i = 1, n * n do
  edge[i] = {}
  srccnt[i] = 0
end
local function getindex(a, b)
  if a < b then
    return (a - 1) * n + b
  else
    return (b - 1) * n + a
  end
end
for i = 1, n do
  local prvidx = getindex(i, io.read("*n"))
  for j = 2, n - 1 do
    local nxtidx = getindex(i, io.read("*n"))
    edge[prvidx][nxtidx] = true
    srccnt[nxtidx] = srccnt[nxtidx] + 1
    prvidx = nxtidx
  end
end
local tasks = {}
for i = 1, n * n do
  if next(edge[i]) and srccnt[i] == 0 then
    table.insert(tasks, i)
  end
end
local day = 0
local matched = 0
while 0 < #tasks do
  day = day + 1
  local ng = false
  local newtasks = {}
  for i = 1, #tasks do
    local src = tasks[i]
    matched = matched + 1
    for dst, _unused in pairs(edge[src]) do
      srccnt[dst] = srccnt[dst] - 1
      if srccnt[dst] == 0 then
        table.insert(newtasks, dst)
      end
    end
    if ng then break end
  end
  if ng then break end
  tasks = newtasks
end
if matched * 2 ~= n * (n - 1) then
  day = -1
end
print(day)