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

// 型定義
typedef long long ll;
typedef pair<ll, ll> P;

// forループ
#define REP(i,n) for(ll i=0; i<(ll)(n); ++i)

// 定数宣言
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

// グラフ表現
using Graph = vector<vector<int>>;

// グラフの辺表現
using Edge = map<pair<int,int>,int>;

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。
template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

// 最大公約数
ll gcd(ll a,ll b){
   if (a%b == 0) return(b);
   else return(gcd(b, a%b));
}

// 最小公倍数
ll lcm(ll a, ll b){
    return a/gcd(a, b) * b;
}

int main()
{
    cout << fixed << setprecision(15);
    ll N;
    cin >> N;

    vector<queue<ll>> A(N);

    REP(i, N){
        REP(j, N-1){
            ll a;
            cin >> a;
            A[i].push(a-1);
        }
    }

    ll ans = 0;
    ll total = N*(N-1)/2;
    bool check = true;
    while(total>0 && check){
        vector<bool> B(N, false);

        check = false;
        REP(i, N){
            // もう全ての試合が終わっているとき
            if(A[i].empty()) continue;
            
            // 自分が今日すでに試合済みのとき
            if(B[i]) continue;

            // 相手が今日すでに試合済みのとき
            if(B[A[i].front()]) continue;
            
            // 次の試合相手と両思いのとき
            if(A[A[i].front()].front() == i){
                // 試合を抜き出しで今日を試合済みに
                A[A[i].front()].pop();
                B[A[i].front()] = true;
                A[i].pop();
                B[i] = true;

                // 総試合数のカウントを引く 
                total--;

                // 今日試合が行われたことのチェック
                check = true;
            }
        }
        ans++;
        // 1試合も行えなかったとき
        if(check == false){
            cout << -1 << endl;
            return 0;
        }
    }
    cout << ans << endl;
    return 0;
}