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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ELeague solver = new ELeague();
        solver.solve(1, in, out);
        out.close();
    }

    static class ELeague {
        public void solve(int testNumber, Scanner in, PrintWriter out) {

            int n = in.nextInt();
            Graph g = new Graph(n * n + 1);
            int[][] match = new int[n][n - 1];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n - 1; j++) {
                    int oponent = in.nextInt();
                    int min = Math.min(i + 1, oponent);
                    int max = Math.max(i + 1, oponent);
                    match[i][j] = min * n + max;
                }
            }

            // 0からほかのやつに辺を伸ばせばいい
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n - 1; j++) {
                    if (j == 0) {
                        g.addEdge(new Edge(0, match[i][j], -1));
                    } else {
                        g.addEdge(new Edge(match[i][j - 1], match[i][j], -1));
                    }
                }
            }

            long[] dist = g.bellmanFord(0);
            if (g.HasNegativeLoop()) {
                out.println(-1);
            } else {
                long min = 1;
                for (int i = 0; i < dist.length; i++) {
                    min = Math.min(min, dist[i]);
                }
                out.println(-min);
            }


        }

    }

    static class Graph {
        int n;
        ArrayList<Edge>[] adj;
        long[] vertexCost;
        boolean hasNegativeLoop = false;

        public Graph(int size) {
            n = size;
            adj = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                adj[i] = new ArrayList<>();
            }
            vertexCost = new long[n];
        }

        public void addEdge(Edge e) {
            adj[e.from].add(e);
        }

        public long[] bellmanFord(int from) {
            hasNegativeLoop = false;
            long[] res = new long[n];
            for (int i = 0; i < n; i++) {
                res[i] = i == from ? 0 : (long) 1e30;
            }

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    for (Edge e : adj[j]) {
                        int f = j;
                        int t = e.to;
                        if (res[t] > res[f] + e.cost) {
                            res[t] = res[f] + e.cost;
                            if (i == n - 1) {
                                hasNegativeLoop = true;
                                break;
                            }
                        }
                    }
                }
            }
            return res;
        }

        public boolean HasNegativeLoop() {
            return this.hasNegativeLoop;
        }

    }

    static class Edge {
        int from;
        int to;
        long cost;

        public Edge(int to) {
            this.to = to;
        }

        public Edge(int to, long cost) {
            this.to = to;
            this.cost = cost;
        }

        public Edge(int from, int to, long cost) {
            this.from = from;
            this.to = to;
            this.cost = cost;
        }

        public String toString() {
            return "Edge{" +
                    "from=" + from +
                    ", to=" + to +
                    ", cost=" + cost +
                    '}';
        }

    }
}