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: Time Limit Exceeded)


import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();

        final Map<Integer, List<Integer>> graph = new HashMap<>();
        final Set<Integer> startVertex = new HashSet<>();
        for (int i = 1; i <= n; i++) {
            int parent = makeVertex(i, scanner.nextInt(), n);
            startVertex.add(parent);
            for (int j = 2; j <= n - 1; j++) {
                final int current = makeVertex(i, scanner.nextInt(), n);
                startVertex.remove(current);
                graph.computeIfAbsent(parent, v -> new ArrayList<>()).add(current);
                parent = current;
            }
        }

        int max = -1;
        for (final int vertex : startVertex) {
            final int depth = dfs(graph, vertex, new boolean[2_000_000]);
            if (depth == -1) {
                System.out.println(-1);
                return;
            }
            max = Math.max(max, depth);
        }
        System.out.println(max);
    }

    private static int dfs(final Map<Integer, List<Integer>> graph, final int current, final boolean[] isVisited) {
        isVisited[current] = true;
        final List<Integer> children = graph.get(current);
        if (children == null) {
            return 1;
        }

        int max = -1;
        for (final int child : children) {
            if (isVisited[child]) {
                return -1;
            }
            final int depth = dfs(graph, child, Arrays.copyOf(isVisited, 2_000_000));
            if (depth == -1) {
                return -1;
            }
            max = Math.max(max, depth);
        }
        return max + 1;
    }

    private static int makeVertex(final int a, final int b, final int n) {
        return Math.min(a, b) * (n + 1) + Math.max(a, b);
    }
}