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: Accepted)

import math, sys
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from copy import deepcopy
from functools import lru_cache
from heapq import heapify, heappop, heappush
from itertools import accumulate, combinations, permutations
input = sys.stdin.readline
mod = 10**9 + 7
ns = lambda: input().strip()
ni = lambda: int(input().strip())
nm = lambda: map(int, input().split())
nl = lambda: list(map(int, input().split()))
sys.setrecursionlimit(10**6)

def main():
    maxn = 1005
    maxv =  maxn * (maxn-1) // 2
    to = [[] for _ in range(maxv)] # 頂点間の辺の情報
    id = [[[] for _ in range(maxn)] for _ in range(maxn)] # 試合のid=DAGの頂点番号
    
    def toId(i,j): # 選手idから試合idを返す関数
        if i > j: # i,jが逆になっても試合idは同じ
            i,j = j,i
        return id[i][j]

    visited = [False]*maxv
    calculated = [False]*maxv
    dp = [1]*maxv # Vからスタートしたときの最長経路。頂点の個数ベースで経路の長さを数えるので、初期値は1

    def dfs(v):
        if visited[v]:
            if not calculated[v]:
                return -1 # 計算が終わっていない頂点を2度訪れるのはループがあるということ
            return dp[v]
        visited[v] = True
        for u in to[v]: # 全ての辺をなめる
            res = dfs(u)
            if res == -1: return -1 # ループがあれば-1を返す
            dp[v] = max(dp[v], res+1)
        calculated[v] = True
        return dp[v]

    n = int(input())
    a = [nl() for _ in range(n)]
    for i in range(n):
        for j in range(n-1):
            a[i][j] -= 1 # 選手idを0始まりに変換
    V = 0
    for i in range(n):
        for j in range(n):
            if i < j:
                id[i][j] = V
                V += 1 # 0から順に各試合にidを割り振る
    for i in range(n):
        for j in range(n-1):
            a[i][j] = toId(i,a[i][j]) # 選手idを試合id(頂点番号)に置き換える
        for j in range(n-2): # 頂点間の依存関係はn-2個
            to[a[i][j+1]].append(a[i][j])
    ans = 0
    for i in range(V):
        res = dfs(i)
        if res == -1:
            print('-1') # ループがあれば-1を返す（問題文の指示）
            return
        ans = max(ans, res)
    print(ans)
    

if __name__ == '__main__':
    main()