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;

typedef long long ll;

int main(){
	ll w[11][11] = {};
	w[0][1] = w[1][0] = 1;
	w[2][0] = w[0][2] = 2;
	w[2][1] = w[1][2] = 3;
	w[3][0] = w[0][3] = 4;
	w[3][1] = w[1][3] = 8;
	w[3][2] = w[2][3] = 13;
	ll l = 23;
	for(int k = 4 ; k < 10 ; k ++){
		vector<ll> vec;
		vec.push_back(1);
		vec.push_back(2);
		while(vec.size() < k){
			vec.push_back(vec[vec.size()-2]+vec[vec.size()-1]+1);
		}
		int cnt = 0;
		for(int i = 1 ; i <= k-2 ; i ++){
			cnt += i;
		}
		for(int t = 0 ; t < k ; t ++){
			w[k][t] = w[t][k] = (l-cnt)*vec[t];
		}
		l = (l-cnt)*(vec[vec.size()-2]+vec[vec.size()-1]+1)-12;
	}
	int n;
	cin >> n;
	for(int i = 0 ; i < n ; i ++){
		for(int j = 0 ; j < n ; j ++){
			printf("%lld%c",w[i][j],(j==n-1)?'\n':' ');
		}
	}
}