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)

/** 
 * Dont raise your voice, improve your argument.
 * --Desmond Tutu
 */

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

typedef long long ll;
//#define endl "\n"
#define fori(n) for(ll i=0; i<(n); i++)

#define cins(s) string s; cin>>s;
#define cini(i) int i; cin>>i;
#define cinll(l) ll l; cin>>l;
#define cinai(a, n) vi a(n); fori(n) { cin>>a[i]; }
#define cinall(a, n) vll a(n); fori(n) { cin>>a[i]; }
#define cinas(s, n) vs s(n); fori(n) { cin>>s[i]; }

typedef pair<int, int> pii;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<string> vs;
typedef vector<ll> vll;

int main() {
cin.tie(nullptr);
std::ios::sync_with_stdio(false);
    cini(n);

    vvi a(n, vi(n));

    vb today(n);
    //set<int> today; // today[i] set of players playing today
    int day=1;

    for(int i=0; i<n; i++) {
        for(int j=0; j<n-1; j++) {
            cin>>a[i][j];
            a[i][j]--;
        }
    }

    int cnt=n*(n-1)/2;
    vi next(n); // a[i][next[i]] ==next opp of player i

/* return true if p1 and p2 can play today against each other.
 */
    function<bool(int,int)> playToday=[&](int p1, int p2) {
        if(a[p1][next[p1]]==p2 && a[p2][next[p2]]==p1) {
            if(today[p1] || today[p2]) {
                return false;
            }
            today[p1]=true;
            today[p2]=true;
            return true;
        }
        return false;
    };

/* return true if p1 and p2 can play against as next, 
 * increments today if neccessary.
 */
    function<bool(int,int)> play=[&](int p1, int p2) {
//cout<<"play p1="<<p1<<" p2="<<p2<<endl;
        if(a[p1][next[p1]]==p2 && a[p2][next[p2]]==p1) {
            if(today[p1] || today[p2]) {
                day++;
                today.assign(n, false);
            }
            today[p1]=true;
            today[p2]=true;
//cout<<"ret true"<<endl;
            return true;
        }
//cout<<"ret false"<<endl;
        return false;
    };

    int minI=0;
    while(cnt) {
        for(int i=minI; i<n; i++) {
            if(next[i]==n-1) {
                if(minI==i)
                    minI++;
                continue;
            }

            int opp=a[i][next[i]];
            if(playToday(i, opp)) {
                next[i]++;
                next[opp]++;
                cnt--;
            }
        }

        bool ok=false;
        for(int i=minI; i<n; i++) {
            if(next[i]==n-1)
                continue;

            int opp=a[i][next[i]];
            if(play(i, opp)) {
                next[i]++;
                next[opp]++;
                ok=true;
                cnt--;
                break;
            }
        }
        if(!ok && cnt) {
            cout<<"-1"<<endl;
            return 0;
        }
    }

    cout<<day<<endl;
}