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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000
#define MOD 1000000007
#define all(v) v.begin(),v.end()
#define pb push_back
int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int kaijo[200010];
struct edge{int to,cost;};
int gcd(int a,int b){
	if(b==0)return a;
	return gcd(b,a%b);
}
int lcm(int a,int b){
	return a/gcd(a,b)*b;
}
bool prime(int a){
	if(a==1)return false;
	for(int i=2;i*i<=a;i++){
		if(a%i==0)return false;
	}
	return true;
}
bool kosa(double ax,double ay,double bx,double by,double cx,double cy,double dx,double dy){//線分の交差判定
	double ta=(cx-dx)*(ay-cy)+(cy-dy)*(cx-ax);
	double tb=(cx-dx)*(by-cy)+(cy-dy)*(cx-bx);
	double tc=(ax-bx)*(cy-ay)+(ay-by)*(ax-cx);
	double td=(ax-bx)*(dy-ay)+(ay-by)*(ax-dx);
	return tc*td<0&&ta*tb<0;
}
void init_fact(int n){
	kaijo[0]=1;
	for(int i=1;i<=n;i++){
		kaijo[i]=kaijo[i-1]*i;
		kaijo[i]%=MOD;
	}
}
int modpow(int a,int b){
	if(b==0)return 1;
	if(b%2)return modpow(a,b-1)*a%MOD;
	int memo=modpow(a,b/2);
	return memo*memo%MOD;
}
int comb(int a,int b){
	if(b>a)return 0;
	if(a==b)return 1;
	return kaijo[a]*modpow(kaijo[a-b],MOD-2)%MOD*modpow(kaijo[b],MOD-2)%MOD;
}
int g(int x){
	x=modpow(x,MOD-2);
	return x;
}
int n,a[1010][1010],day=0;
bool b[1010][1010];
P now[1010];
//次の相手、最終の試合の日
signed main(){
	cin>>n;
	rep(i,n){
		rep(j,n-1){
			cin>>a[i][j];
			a[i][j]--;
		}
	}
	rep(i,n)now[i]={0,-1};
	queue<int>q;
	while(1){
		bool flag=false;
		rep(i,n){
			if(now[i].fi==n-1)continue;
			int aite=a[i][now[i].fi];
			if(now[aite].fi==n-1)continue;
			if(a[aite][now[aite].fi]==i){
				if(b[i][aite])continue;
				if(now[i].se==day||now[aite].se==day)continue;
				b[i][aite]=b[aite][i]=true;
				
				now[i].fi++,now[i].se=day;
				now[aite].fi++,now[aite].se=day;
				flag=true;
			}
		}
		if(!flag)break;
		day++;
	}
	rep(i,n){
		rep(j,n){
			if(i!=j){
				if(!b[i][j]){
					puts("-1");
					return 0;
				}
			}
		}
	}
	cout<<day<<endl;
	return 0;
}