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;
typedef signed long long ll;
 
#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------
 
int N;
ll W[10][10];
 
void solve() {
	int i,j,k,l,r,x,y; string s;
	
	W[1][0]=W[0][1]=1;
	for(N=3;N<=10;N++) {
		vector<int> V;
		FOR(i,N-1) V.push_back(i);
		ll ma=0;
		do {
			if(V[0]>V.back()) continue;
			ll ret=0;
			FOR(i,N-2) ret+=W[V[i]][V[i+1]];
			ma=max(ma,ret);
		} while(next_permutation(ALL(V)));
		
		ma++;
		int cand[]={1,2,4,7,12,20,29,38,52,73};
		FOR(x,N-1) {
			W[N-1][x]=W[x][N-1]=ma*cand[x];
		}
	}
	
	cin>>N;
	vector<int> V;
	FOR(i,N) V.push_back(i);
	set<ll> C;
	
	do {
		if(V[0]>V.back()) continue;
		ll ret=0;
		FOR(i,N-1) ret+=W[V[i]][V[i+1]];
		assert(C.count(ret)==0);
		
	} while(next_permutation(ALL(V)));
	
	
	FOR(y,N) {
		FOR(x,N) cout<<W[y][x]<<" ";
		cout<<endl;
	}
}
 
 
int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}