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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int w[11] = {0, 1, 2, 4, 7, 12, 20, 29, 38, 52, 73};

template <typename Tp> inline void getint(Tp &num){
	register int ch, neg = 0;
	while(!isdigit(ch = getchar())) if(ch == '-') neg = 1;
	num = ch & 15;
	while(isdigit(ch = getchar())) num = num * 10 + (ch & 15);
	if(neg) num = -num;
}

int N, p[11];
ll ed[11][11], M = 0;

int main(){
	getint(N);
	for(register int i = 2; i <= N; i++){
		for(register int j = 1; j < i; j++) ed[i][j] = ed[j][i] = (M + 1) * w[j];
		if(i == N) break;
		for(register int j = 1; j <= i; j++) p[j] = j;
		do {
			register ll len = 0;
			for(register int j = 1; j < i; j++) len += ed[p[j]][p[j + 1]];
			M = max(M, len);
		} while(next_permutation(p + 1, p + i + 1));
	}
	for(register int i = 1; i <= N; i++, puts(""))
		for(register int j = 1; j <= N; j++) printf("%lld ", ed[i][j]);
	return 0;
}