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)

from collections import deque
import sys
input = sys.stdin.readline

N = int(input().strip("\n"))
A = [list(map(lambda x :int(x)-1, input().split())) for _ in range(N)]

# 対戦カード (インデックス0>インデックス[1]だけ使う)
all_patterns = [[[] for _ in range(N)] for _ in range(N)]
lines_to_here = [[0 for _ in range(N)] for _ in range(N)]

for i in range(len(A)):
    for j in range(len(A[i])-1):
        playerA = i
        playerB = A[i][j]
        if playerA < playerB:
            playerA, playerB = playerB, playerA
        next_playerA = i
        next_playerB = A[i][j+1]
        if next_playerA < next_playerB:
            next_playerA, next_playerB = next_playerB, next_playerA
        
        all_patterns[playerA][playerB].append([next_playerA,next_playerB])
        lines_to_here[next_playerA][next_playerB] += 1

next_out = deque()
for i in range(1,N):
    for j in range(i):
        if lines_to_here[i][j] == 0:
            next_out.append([i,j])

if len(next_out) == 0:
    print(-1)
    sys.exit()

indexin_topological_sorted = [[0 for _ in range(N)] for _ in range(N)]

# topological sort
topological_sorted = []
while len(next_out) > 0:
    added = next_out.pop()
    topological_sorted.append(added)
    indexin_topological_sorted[added[0]][added[1]] = len(topological_sorted)-1
    for direction in all_patterns[added[0]][added[1]]:
        lines_to_here[direction[0]][direction[1]]-=1
        if lines_to_here[direction[0]][direction[1]] == 0:
            next_out.append(direction)

# if there is closed circuit, number remain(not 0 exists)
for i in range(1,N):
    for j in range(i):
        if lines_to_here[i][j] > 0:
            print(-1)
            sys.exit()

costs = [0 for _ in range(len(topological_sorted))]

max_cost = 0
for i in range(len(topological_sorted)):
    processing = topological_sorted[i]
    now_cost = costs[i]
    for direction in all_patterns[processing[0]][processing[1]]:
        direction_in_topological_sorted = indexin_topological_sorted[direction[0]][direction[1]]        
        costs[direction_in_topological_sorted] = max(1+now_cost,costs[direction_in_topological_sorted])
        max_cost = max(max_cost,costs[direction_in_topological_sorted])
print(max_cost+1)