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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
long double ABS(long double a){return max(a,-a);}
long long g[20][20];
int perm[20];
int mag[]={1,2,4,7,12,20,29,38,52,73};
int main(){
	int a;scanf("%d",&a);
	g[0][1]=g[1][0]=1;
	for(int i=3;i<=a;i++){
		for(int j=0;j<i-1;j++)perm[j]=j;
		long long cur=0;
		do{
			long long tmp=0;
			for(int j=0;j<i-2;j++){
				tmp+=g[perm[j]][perm[j+1]];
			}
			cur=max(cur,tmp);
		}while(next_permutation(perm,perm+i-1));
		cur++;
		for(int j=0;j<i-1;j++){
	//		g[j][i-1]=g[i-1][j]=mag[i-2-j]*cur;
		g[j][i-1]=g[i-1][j]=mag[j]*cur;
		}
	}
	for(int i=0;i<a;i++){
		for(int j=0;j<a;j++){
			if(j)printf(" ");printf("%lld",g[i][j]);
		}
		printf("\n");
	}
}