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: Runtime Error)

#include<bits/stdc++.h>
#include<string>
#define pii pair<int,int>
#define fr(j,n) for(int j=0;j<n;j++)
#define f(n) for(int i=0;i<n;i++)
#define rt return 0;
#define sc second
#define fi first
#define i(n) int n;cin>>n;
#define pb(i) push_back(i);
#define rt return 0;
#define sz(v) v.size()
#define allv v.begin(),v.end()
#define ff(i,in,fi) for(int i=in;i<fi;i++)
//#define df int n;cin>>n;int a[n];fr(n){cin>>a[i];}5 8 7 14 55 19 9 2
using namespace std;
#define trav(a, x) for(auto& a : x)
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl
#define invect(data,n,commands) for(int xxx = 0;xxx<n;xxx++){int tmp;cin>>tmp;data.pb(tmp);commands}
#define ll long long int
#define p(s) cout<<#s<<":"<<s<<'\n';
#define here(n) cout<<"here"<<n<<endl;
#define ios         ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');std::cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
std::vector<int> v[1006];
int n;
int check()
{
	ff(i,1,n+1)
	{
		if(sz(v[i]))
		{
			return 1;
		}
	}
	return 0;
}
int main()
{
	ios;
	cin>>n;
	f(n)
	{
		int a;
		fr(j,n-1)
		{
			cin>>a;
			v[i+1].pb(a);
		}
	}
	int days=0;
	int vis[n+1];
	while(check())
	{
		f(n+1)
		vis[i]=0;
		int flag=0;
		ff(i,1,n+1)
		{
			if(i==v[v[i][0]][0]&&vis[i]==0&&vis[v[i][0]]==0)
			{
				flag=1;
				here(11);			
				vis[i]=1;
				vis[v[i][0]]=1;
				v[v[i][0]].erase(v[v[i][0]].begin());
				v[i].erase(v[i].begin());
			}
		}
		//trace(vis[3],vis[4]);
		/*ff(i,1,n+1)
		{
			trav(a,v[i])
			{
				trace(a,i);
			}
		}*/
		days++;
		if(flag==0)
		{
			//trace(days);
			cout<<-1;
			rt;
		}
	}
	cout<<days;
}