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: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <tuple>
#include <utility>
#include <vector>
#define ll long long
#define itn int
#define endl '\n'
#define co(ans) cout<<ans<<endl
#define COYE cout<<"YES"<<endl
#define COYe cout<<"Yes"<<endl
#define COye cout<<"yes"<<endl
#define CONO cout<<"NO"<<endl
#define CONo cout<<"No"<<endl
#define COno cout<<"no"<<endl
#define FORE(i,a) for(auto &i:a)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RREP(i,n)  FFOR(i,1,n)
#define PB push_back
#define MP make_pair
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort((V).begin(),(V).end())
#define REVERSE(V) reverse((V).begin(),(V).end())
#define EACH(V,i) for(typeof((V).begin()) i=(V).begin();i!=(V).end();++i)
#define equals(a,b) (fabs((a)-(b))<EPS)
#define INF ((1LL<<62)-(1LL<<31))
#define EPS 1e-10
#define PI 3.141592653589793238
#define MOD 1000000007
#define MAX 5100000
using namespace std;
using Edge=pair<int,int>;
using Graph=vector<vector<int>>;
inline int toInt(string s){int v;istringstream sin(s);sin>>v;return v;}
template<class T>inline string toString(T x){ostringstream sout;sout<<x;return sout.str();}
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;}
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int N;
deque<int> A[1010];
int step[1010];
int M;
int main(){
  cin>>N;
  REP(i,N){
    REP(j,N-1){
      int x;
	  cin>>x;
	  A[i].push_back(x-1);
    }
  }
  M=N*(N-1)/2;
  
  queue<int> Q;
  REP(i,N) Q.push(i);
  
  int num=0;
  while(Q.size()){
    queue<int> Q2;
    
    while(Q.size()){
      int i=Q.front();
      Q.pop();
      if(step[i]<=num && A[i].size() && A[A[i].front()].front()==i && step[A[i].front()]<=num) {
        step[i]=step[A[i].front()]=num+1;
        Q2.push(i);
        Q2.push(A[i].front());
        A[A[i].front()].pop_front();
        A[i].pop_front();
        M--;
      }
    }
    num++;
    if(M==0) break;
    swap(Q,Q2);
  }
  
  if(M) cout<<-1<<endl;
  else cout<<num<<endl;
}