Problem:
Score : 900 points

Problem Statement

Diverta City is a new city consisting of N towns numbered 1, 2, ..., N.

The mayor Ringo is planning to connect every pair of two different towns with a bidirectional road. The length of each road is undecided.

A Hamiltonian path is a path that starts at one of the towns and visits each of the other towns exactly once.
The reversal of a Hamiltonian path is considered the same as the original Hamiltonian path.

There are N! / 2 Hamiltonian paths. Ringo wants all these paths to have distinct total lengths (the sum of the lengths of the roads on a path), to make the city diverse.

Find one such set of the lengths of the roads, under the following conditions:

The length of each road must be a positive integer.

The maximum total length of a Hamiltonian path must be at most 10^{11}.

Constraints

N is a integer between 2 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print a set of the lengths of the roads that meets the objective, in the following format:

w_{1, 1} \ w_{1, 2} \ w_{1, 3} \ ... \ w_{1, N}
w_{2, 1} \ w_{2, 2} \ w_{2, 3} \ ... \ w_{2, N}
: : :
w_{N, 1} \ w_{N, 2} \ w_{N, 3} \ ... \ w_{N, N}

where w_{i, j} is the length of the road connecting Town i and Town j, which must satisfy the following conditions:

w_{i, i} = 0

w_{i, j} = w_{j, i} \ (i \neq j)

1 \leq w_{i, j} \leq 10^{11} \ (i \neq j)

If there are multiple sets of lengths of the roads that meet the objective, any of them will be accepted.

Sample Input 1

3

Sample Output 1

0 6 15
6 0 21
15 21 0

There are three Hamiltonian paths. The total lengths of these paths are as follows:

1 → 2 → 3: The total length is 6 + 21 = 27.

1 → 3 → 2: The total length is 15 + 21 = 36.

2 → 1 → 3: The total length is 6 + 15 = 21.

They are distinct, so the objective is met.

Sample Input 2

4

Sample Output 2

0 111 157 193
111 0 224 239
157 224 0 258
193 239 258 0

There are 12 Hamiltonian paths, with distinct total lengths.

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

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args){
        PrintWriter out = new PrintWriter(System.out);
        FastScanner sc = new FastScanner();

        int N = sc.nextInt();

        for (int i=0;i<N;i++) {
            for (int j=0;j<N;j++) {
                long l = 0;

                if (i != j) {
                    l |= ((long) j) << (3 * i);
                    l |= ((long) i) << (3 * j);
                }

                out.printf("%d", l);

                if (j < N-1) {
                    out.printf(" ");
                }
            }
            out.println();
        }

        out.flush();
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(String s) {
            try {
                br = new BufferedReader(new FileReader(s));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String nextToken() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(nextToken());
        }

        long nextLong() {
            return Long.parseLong(nextToken());
        }

        double nextDouble() {
            return Double.parseDouble(nextToken());
        }

        int[] nextIntArray(int N) {
            int a[] = new int[N];
            for (int i=0;i<N;i++) {
                a[i] = this.nextInt();
            }

            return a;
        }

        long[] nextLongArray(int N) {
            long a[] = new long[N];
            for (int i=0;i<N;i++) {
                a[i] = this.nextLong();
            }

            return a;
        }

        double[] nextDoubleArray(int N) {
            double a[] = new double[N];
            for (int i=0;i<N;i++) {
                a[i] = this.nextDouble();
            }

            return a;
        }

        String[] nextTokenArray(int N) {
            String a[] = new String[N];
            for (int i=0;i<N;i++) {
                a[i] = this.nextToken();
            }

            return a;
        }
    }
}