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`.

Examples

Input

3
2 3
1 3
1 2


Output

3


Input

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


Output

4


Input

3
2 3
3 1
1 2


Output

-1

from collections import deque

N = int(input())
match_order = [deque(map(lambda x: int(x) - 1, input().split())) for _ in range(N)]


q = deque(range(N))
finish_by = [0] * N
next_opponent = [None] * N

while q:
    a = q.popleft()
    # aの残りの順で一番初めに戦う対戦相手を取り出す
    b = match_order[a].popleft()
    if next_opponent[b] == a:
        finish_by[a] = finish_by[b] = max(finish_by[a], finish_by[b]) + 1
        # a,bについて、まだ対戦相手が残っているならキューに積む
        if match_order[a]:
            q.append(a)
        if match_order[b]:
            q.append(b)
    # bの残りの順で戦う相手がaではないなら、順番が来るまで待つ
    else:
        next_opponent[a] = b

# 対戦相手が残っている人がいる場合、条件通りに終わることができない
if any(match_order):
    print(-1)
else:
    print(max(finish_by))