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>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
typedef long long LL;
const int f[]={0,1,2,4,7,12,20,29,38,52,73};
int n,now;LL mx,w[11][11];bool vis[11];
void dfs(int k1,LL k2,int k3){
	if(k1>now){
		if(k2>mx)mx=k2;
		return;
	}
	rep(i,1,n)if(!vis[i]&&w[k1][i]){
		vis[i]=1;
		dfs(k1+1,k2+w[k3][i],i);
		vis[i]=0;
	}
}
int main(){
	scanf("%d",&n);
	rep(i,1,n){
		rep(j,1,i-1)w[i][j]=w[j][i]=1LL*f[j]*(mx+1);
		now=i;
		rep(j,1,i)dfs(j,0,0);
	}
	rep(i,1,n){rep(j,1,n)printf("%lld ",w[i][j]),assert(w[i][j]<=1e12);puts("");}
	return 0;
}