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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
#define ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define uniq(v) v.erase(unique(v.begin(),v.end()),v.end())
#define lb(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define ub(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
using Pair = pair<ll,pair<int,int>>;
#define pq priority_queue<Pair, vector<Pair>, greater<Pair>> 
const ll mod=1000000007;
//const ll mod=998244353;
const ld pi=acos(-1.0);
const ll INF = 1LL<<61;
template<class T>bool chmax(T &a, const T &b) { 
  if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) {
  if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'大文字'+=32;    で小文字に
//  cout << fixed << setprecision (20);   小数点以下2０桁まで
//実行時間制約2秒では２×10^8回くらいまで計算できる



int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin>>n;

  vector<queue<int>> p(n);
  rep(i,n){
    rep(j,n-1){
      int x;
      cin>>x;
      x--;
      p[i].push(x);
    }
  }

  int ans=0;
  while(1){
    bool dame=true;
    bool kara=true;

    ans++;

    vector<bool> che(n,false);
    rep(i,n){
      if(!che[i] && !p[i].empty()){
        kara=false;
        int x=p[i].front();
        if(p[x].empty()) continue;
        int y=p[x].front();
        if(i==y){
          dame=false;
          che[x]=true;
          p[i].pop();
          p[x].pop();
        }
      }
    }
    if(kara) {
      cout << ans << endl;
      return 0;
    }
    if(dame) {
      cout << -1 << endl;
      return 0;
    }
  }

}