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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define fw(p) for(int w=0;w<(p);w++)
#define fx(p) for(int x=0;x<(p);x++)
#define fy(p) for(int y=0;y<(p);y++)
#define fz(p) for(int z=0;z<(p);z++)
#define fyg(p,g) for(int y=(g);y<(p);y++)
#define fzg(p,g) for(int z=(g);z<(p);z++)
#define ce(d) cout<<d<<endl;
#define vecp(p) int aa;cin>>aa;(p).push_back(aa);
#define vecpl(p) long long aa;cin>>aa;(p).push_back(aa);
#define vecps(p) string aa;cin>>aa;(p).push_back(aa);
#define vecp2(p) cin>>aa;(p).push_back(aa);
#define vecpl2(p) long long a	b;cin>>ab;(p).push_back(ab);
#define vecps2(p) string ab;cin>>ab;(p).push_back(ab);
#define set0(k,n) for(int nn=0;nn<(n);nn++){ (k).push_back(0); }
#define sorts(c) sort((c).begin(),(c).end());
#define reverses(c) reverse((c).begin(),(c).end());
#define vec(b) vector<int> (b);
#define vecl(b) vector<long long> (b);
#define vecs(b) vector<string> (b);
#define vecsize(b,size) vector<int> (b)((size));
#define pb(b,a) (b).push_back((a));
#define doublece(a,b) cout<<(a)<<' '<<(b)<<endl;
#define pairs(s) vector<pair<int,int>> (s);
#define pairsl(s) vector<pair<ll,ll>> (s);
#define pairss(s) vector<pair<string,string>> (s);
#define pairsp(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairspl(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsps(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsREV(s) (s).push_back(make_pair(bb,aa));
#define pairslREV(s) (s).push_back(make_pair(bb,aa));
#define pairssREV(s) (s).push_back(make_pair(bb,aa));
#define MOD 1000000007
int L[1100][1100];

int main()
{
	int N;
	cin >> N;
	vector<vector<int>> A(N + 1);
	fx(N) {
		fy(N-1) {
			int a;
			cin >> a;
			pb(A.at(x+1),a)
		}
		L[x + 1][A.at(x + 1).at(0)] = 1;
	}
	int ans = 0;
	while (true){
		ans++;
		bool flag = true;
		bool b[1100];
		fx(N) {
			b[x + 1] = false;
		}
		fx(N) {
			if (A.at(x + 1).size() == 0) { continue; }
			if (b[x + 1] || ( b[A.at(x + 1).at(0)])) { continue; }
			if (L[x+1][A.at(x+1).at(0)]>0&& L[A.at(x + 1).at(0)][x+1]>0) {
				int k = A.at(x + 1).at(0);
				A.at(x + 1).erase(A.at(x+1).begin());
				A.at(k).erase(A.at(k).begin());
				b[x + 1] = true;
				b[k] = true;
				if (A.at(x + 1).size()>0) {
					L[x + 1][A.at(x + 1).at(0)] = 1;
				}
				if (A.at(k).size()>0) {
					L[k][A.at(k).at(0)] = 1;
				}

				flag = false;
			}
		}

		if (flag) {
			break;
		}

	}
	fx(N) {
		if (A.at(x+1).size()>0) {
			ce(-1)
				return 0;
		}
	}
	ce(ans-1)
    return 0;
}