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;
#define int long long

int max_ham(vector<vector<int>> &w, int N){
	vector<int> p(N);
	int i, ans = 0;
	for(i = 0; i < N; i++){
		p[i] = i;
	}
	do{
/*		for(i = 0; i < N; i++){
			printf("%lld ", p[i]);
		}
		printf("\n");
*/		int now = 0;
		for(i = 1; i < N; i++){
			now += w[p[i - 1]][p[i]];
		}
		ans = max(ans, now);
	}while(next_permutation(p.begin(), p.end()));
	return ans;
}

signed main(){
	int N, i, j;
	int a[] = {1, 2, 4, 7, 12, 20, 29, 38, 52, 73};
	scanf("%lld", &N);
	vector<vector<int>> w(N, vector<int>(N, 0));
	w[0][1] = 1;
	w[1][0] = 1;
	for(i = 2; i < N; i++){
//		printf("i = %lld\n", i);
		int M = max_ham(w, i);
//		printf("M = %lld\n", M);
		for(j = 0; j < i; j++){
			w[i][j] = a[j] * (M + 1);
			w[j][i] = a[j] * (M + 1);
		}
	}
//	printf("%lld\n", max_ham(w, N));
	for(i = 0; i < N; i++){
		for(j = 0; j < N; j++){
			printf("%lld ", w[i][j]);
		}
		printf("\n");
	}
	return 0;
}