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>
#include <algorithm>
#include <set>
#define llint long long

using namespace std;

llint n;
llint a[15][15];
llint f[105];

int main(void)
{
	cin >> n;
	a[1][2] = a[2][1] = 1;
	
	f[0] = 0, f[1] = 1, f[2] = 2;
	for(int z = 3; z <= 10; z++){
		f[z] = f[z-1];
		while(1){
			vector<llint> vec;
			for(int i = 1; i <= z; i++){
				vec.push_back(f[i]);
				for(int j = 1; j <= z; j++){
					if(i < j) vec.push_back(f[i]+f[j]);
				}
			}
			sort(vec.begin(), vec.end());
			bool flag = true;
			for(int i = 1; i < vec.size(); i++){
				if(vec[i-1] == vec[i]) flag = false;
			}
			if(flag) break;
			f[z]++;
		}
	}
	
	for(int i = 3; i <= n; i++){
		llint perm[10], m = 0;
		for(int j = 1; j <= i-1; j++) perm[j] = j;
		do{
			llint sum = 0;
			for(int j = 1; j < i-1; j++) sum += a[perm[j]][perm[j+1]];
			m = max(m, sum);
		}while(next_permutation(perm+1, perm+i));
		
		for(int j = 1; j < i; j++){
			a[i][j] = a[j][i] = (m+1)*f[j];
		}
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
	
	return 0;
}