Problem: 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.

Examples

Input

3


Output

0 6 15
6 0 21
15 21 0


Input

4


Output

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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define maxn 20
using namespace std;
ll mp[maxn][maxn], val[maxn], mx, cnt, used[maxn], ch[maxn];
void get_mx(int dep, int n, ll x)
{
	if (dep > n) mx = max(mx, x);
	else
	for (int i = 1; i <= n; i++)
		if (used[i] == 0)
		{
			ch[dep] = i;
			used[i] = 1;
			if (dep == 1) get_mx(dep + 1, n, x);
			else get_mx(dep + 1, n, mp[ch[dep - 1]][i] + x);
			used[i] = 0;
		}
}
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= 1000; i++)
	{
		int flag = 1;
		for (int j = 1; j <= cnt; j++)
			for (int k = j + 1; k <= cnt; k++)
			{
				if (j == k) continue;
				if (val[j] + val[k] == i) flag = 0;
				for (int l = 1; l <= cnt; l++)
					if (val[j] + val[k] == i + val[l]) flag = 0;
			}
		if (flag)
		{
			val[++cnt] = i;
			if (cnt == n) break;
		}
	}
	mp[1][2] = mp[2][1] = 1;
	for (int i = 3; i <= n; i++)
	{
		mx = 0;
		get_mx(1, i - 1, 0);
		for (int j = 1; j <= i - 1; j++)
			mp[j][i] = mp[i][j] = (mx + 1) * val[j];
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
			cout << mp[i][j] << " ";
		cout << endl;
	}
}