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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

import math, string, itertools, fractions, heapq, collections, re,  array, bisect, sys, random, time, copy, functools, os, pdb
sys.setrecursionlimit(10**7)
from fractions import Fraction
inf = 10 ** 20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
ddn = [(-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def _I(): return int(sys.stdin.readline())
def _F(): return float(sys.stdin.readline())
def _pf(s): return print(s, flush=True)
def perr(s): return print(s, file=sys.stderr)

def main():
    A_list = []
    if 'LOCAL' in os.environ:
        N = 1000
        for i in range(N):
            tmp = [j for j in range(N) if i != j]
            random.shuffle(tmp)
            A_list.append(tmp)

    else:
        N = _I()
        for i in range(N):
            A_list.append(LI())
        #  perr(A_list)
    new_A_list = {}
    for aidx, A in enumerate(A_list):
        new_A_list[aidx] = [i-1 for i in A]
    A_list = new_A_list

    """
    まず、N=1000までなので、
    1000*1000=10**6まである。
    試合の日数を最小かする。
    つまりできるだけ効率よく1日に試合を詰め込むことを考える
    そもそも1のところに2とあったら、
    全通りを求めると？
    どの試合にするかで1000*999/2の階乗=10**6/2の階乗通りある
    まず選手iの1つ目を見ていく
    その相手のところをみて、current_idxがiだったら試合可能

    """

    current_opponent_idx = None
    current_day = 0
    is_ok = False
    # 各選手の試合状況
    current_position_list = {i:0 for i in range(N)}
    done_player_list = set()
    position_sum = 0
    done_sum = (N-1)*N
    all_done_players = set()
    while True:
        #  print('day: ', current_day)
        #  print('current position list', current_position_list)
        #  pdb.set_trace()
        is_ok = False
        #  if all(position == N-1 for position in current_position_list):
        if position_sum == done_sum:
            # 全試合終了
            break

        for i in range(N):
            #  print('--- in for---- current position list', current_position_list)
            # 選手iが試合できるか見る
            if i in done_player_list:
                continue

            if i in all_done_players:
                # この選手はすべての試合を終了している
                continue

            # iの試合相手のidx
            opponent_idx = A_list[i][current_position_list[i]]
            if opponent_idx in done_player_list:
                # 試合相手がすでに今日試合してた場合
                continue

            #  print('i', i)
            #  print('opponent', opponent_idx)

            # iの試合相手の次の相手が選手iなら成立する
            if A_list[opponent_idx][current_position_list[opponent_idx]] == i:
                #  print('試合した', i, opponent_idx)
                # 試合可能
                #  done_player_list.extend([i, opponent_idx])
                done_player_list.add(opponent_idx)
                done_player_list.add(i)
                current_position_list[i] += 1
                current_position_list[opponent_idx] += 1
                if current_position_list[i] == N-1:
                    all_done_players.add(i)
                if current_position_list[opponent_idx] == N-1:
                    all_done_players.add(opponent_idx)
                position_sum += 2
                # 試合がまだ可能であるというフラグをつける
                is_ok = True

        if not is_ok:
            # 1試合も成立しなかったらストップする
            print(-1)
            exit()

        done_player_list = set()
        current_day += 1

    print(current_day)
if __name__ == '__main__':
    main()