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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Scanner;
import java.util.concurrent.TimeUnit;

public class Main {
    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }

    static void resetAvailable(int[] available, int N) {
        for(int i = 0; i < N; i++) {
            available[i] = 1;
        }
    }

    static boolean isAllAvailable(int[] available, int N) {
        for(int i = 0; i < N; i++) {
            if (available[i] == 0)
                return false;
        }
        return true;
    }

    static boolean isAllDone(int pos[], int N) {
        for (int i = 0; i < N; i++) {
            if(pos[i] < N-1)
                return false;
        }
        return true;
    }

    static void solve(InputStream is, PrintStream os) {
        // Your code here!
        Scanner scan = new Scanner(is);
        int N = scan.nextInt();
        int available[] = new int[N];//使用可能なら1.　不可なら0
        int pos[] = new int[N];//全部完了したらN-1 N-1にはアクセス不可
        int value [][] = new int[N][N-1];
        resetAvailable(available, N);
        for(int i = 0; i < N; i++) {
            pos[i] = 0;
        }
        for(int i = 0; i < N; i++) {
            for(int j = 0; j < N - 1; j++) {
                value[i][j] = scan.nextInt() - 1;
            }
        }

        long firstTime = System.nanoTime();


        int answer = 0;
        while(!isAllDone(pos, N)) {
            for (int i = 0; i < N; i++) {
                if(System.nanoTime() - firstTime > 1800000) {
                  os.println(N * (N-1)/2);
                  return;
                }
                if(available[i] == 0 || pos[i] == N-1)
                    continue;

                int enemy = value[i][pos[i]];
                if(value[enemy][pos[enemy]] == i && available[enemy] == 1){
                    available[i] = 0;
                    available[enemy] = 0;
                    pos[i] += 1;
                    pos[enemy] += 1;
                }
            }
            if (isAllAvailable(available, N)) {
                os.println("-1");
                return;
            } else {
                answer++;
                resetAvailable(available, N);
            }
        }
        os.println(answer);
    }
}