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 pa pair<int,int>
#define CLR(a,x) memset(a,x,sizeof(a))
#define MP make_pair
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef long double ld;
const int maxn=12;

inline char gc(){
	return getchar();
	static const int maxs=1<<16;static char buf[maxs],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxs,stdin),p1==p2)?EOF:*p1++;
}
inline ll rd(){
    ll x=0;char c=gc();bool neg=0;
    while(c<'0'||c>'9'){if(c=='-') neg=1;c=gc();}
    while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+c-'0',c=gc();
    return neg?(~x+1):x;
}

int f[maxn]={0,1,2,3,6,10,17,30,48,79},N,a[maxn];
ll ans[maxn][maxn];
map<ll,int> mp;

int main(){
    //freopen("","r",stdin);
    N=rd();
    ll now=1;
    
    for(int i=1;i<=N;i++){
    	int n=0;
    	for(int j=1;j<=N&&n<N;j++){
    		if(ans[i][j]||i==j) continue;
    		ans[i][j]=ans[j][i]=f[++n]*now;
    	}
    	now*=f[n]+f[n-1]+1;
    }
    for(int i=1;i<=N;i++) for(int j=1;j<=N;j++) printf("%lld%c",ans[i][j]," \n"[j==N]);
    	
	return 0;
    for(int i=1;i<=N;i++) a[i]=i;
    	ll ma=0;
   	do{
   		ll now=0;
   		for(int i=1;i<N;i++) now+=ans[a[i]][a[i+1]];
   			ma=max(ma,now);
   		// for(int i=1;i<=N;i++) printf("%d ",a[i]);
   			// printf(":%lld\n",now);
   		assert(mp[now]<=1);
   		// assert(now<=1e11);
   		mp[now]+=1;
   	}while(next_permutation(a+1,a+N+1));
   	printf("%lld\n",ma);
    return 0;
}