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 <iostream>
#include <vector>
using namespace std;
using ll = long long;

const int N = 10;
ll cost[N][N];

int main() {
	vector<vector<int>> xr = {
	{1},
	{1,2},
	{1,2,4},
	{1,2,4,7},
	{1,2,4,7,12},
	{1,2,4,8,13,18},
	{1,2,4,8,14,19,24},
	{1,2,4,8,15,24,29,34},
	{1,7,10,13,21,26,41,43,45}
	};

	int n;
	cin >> n;
	ll b = 1;
	for (int i = 0; i+1 < n; ++i) {
		int ln = n-1-i;
		for (int j = i+1; j < n; ++j) {
			cost[i][j] = b * xr[ln-1][j-1-i];
			cost[j][i] = cost[i][j];
		}
		if (ln >= 2) b *= (1+xr[ln-1][ln-1]+xr[ln-1][ln-2]);
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			cout << cost[i][j] << ' ';
		}
		cout << '\n';
	}
}