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>
using namespace std;

typedef long long ll;
const int maxn = 1e5 + 5;

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

const int w[2][9] = {1, 2, 3, 5, 8, 13, 21, 30, 39, 1, 2, 4, 7, 12, 20, 29, 38, 52};

int n, p[15];
ll e[15][15];


ll calc(int n)
{
	ll Max = 0;
	for (int i = 0; i < n; ++i) p[i] = i;
	do {
		ll sum = 0;
		for (int i = 0; i < n - 1; ++i) sum += e[p[i]][p[i + 1]];
		Max = max(Max, sum);
	} while (next_permutation(p, p + n));
	return Max;
}

int main()
{
	n = gi();
	ll M = 0;
	for (int i = 1; i < n; ++i) {
		for (int j = 0; j < i; ++j)
			e[i][j] = e[j][i] = w[1][j] * (M + 1);
		M = calc(i + 1);
		//printf("%lld\n", M);
	}
	
	for (int i = 0; i < n; ++i, puts(""))
		for (int j = 0; j < n; ++j)
			printf("%lld ", e[i][j]);
	
	return 0;
}