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 PyPy2 (5.6.0) (judge verdict: Runtime Error)

# coding: utf-8
from itertools import *
import sys
import time

sys.setrecursionlimit(10**6)


#################### library ####################


#################### graph ####################

# graph: { node: [node] }
def is_cyclic(graph):
	for x0 in graph.keys():
		s = set()
		v = [x0]
		while v:
			x = v.pop()
			for y in graph[x]:
				if y not in s:
					s.add(y)
					v.append(y)
		if x0 in s:
			return True
	else:
		return False


#################### naive ####################


#################### process ####################

def read_input():
	N = int(raw_input())
	table = [ map(int, raw_input().split()) for _ in range(N) ]
	for k in range(N):
		for l in range(N-1):
			table[k][l] -= 1
	return table

def make_graph(table):
	N = len(table)
	dics = [ dict(zip(v, count())) for v in table ]
	graph = dict(((2*N-k-1)*k/2+l-k-1, [])
					for k, l in combinations(range(N), 2))
	for k, l in combinations(range(N), 2):
		i1 = dics[k][l]
		if i1 < N-2:
			j1 = table[k][i1+1]
			v = (2*N-k-1)*k/2+j1-k-1 if k < j1 else (2*N-j1-1)/2+k-j1-1
			graph[(2*N-k-1)*k/2+l-k-1].append(v)
		i2 = dics[l][k]
		if i2 < N-2:
			j2 = table[l][i2+1]
			v = (2*N-l-1)*l/2+j2-l-1 if l < j1 else (2*N-j2-1)/2+l-j2-1
			graph[(2*N-k-1)*k/2+l-k-1].append(v)
	return graph

def reverse_graph(graph):
	rev_graph = dict((v, []) for v in graph.keys())
	for v0, vs in graph.items():
		for v1 in vs:
			rev_graph[v1].append(v0)
	return rev_graph

def f():
	table = read_input()
#	table = [ [1,2], [0,2], [0,1] ]
	N = len(table)
	memo = { }
	graph = make_graph(table)
	if is_cyclic(graph):
		return -1
	
	unvisited = set(range(N*(N-1)/2))
	for v in graph.keys():
		if not graph[v]:
			memo[v] = 1
			unvisited.remove(v)
	
	for v in range(N*(N-1)/2):
		if v not in unvisited:
			continue
		stack = [v]
		while stack:
			v = stack.pop()
			unvs = [ v1 for v1 in graph[v] if v1 in unvisited ]
			if not unvs:
				dist = max(memo[v1]+1 for v1 in graph[v])
				memo[v] = dist
				unvisited.remove(v)
			else:
				stack.append(v)
				for v1 in unvs:
					stack.append(v1)
	
	return max(memo.values())


#################### main ####################

print f()