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 ll long long
#define lld long double
#define MOD 1000000007
#define CAP 100000001
#define inf 1000000000000000000LL
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define endl '\n'
#define sz(v) v.size()
#define all(v)    v.begin(),v.end()
#define fast  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;
ll power(ll x,ll y, ll md=MOD){ll res = 1;x%=md;while(y>0){if(y&1)res = (res*x)%md;x = (x*x)%md;y = y>>1;}return res%md;}
#define dbg(x) cout <<#x<<":"<<x<<endl;
ll max1(ll a, ll b, ll c=-inf, ll d=-inf) {ll mx1=(a>b)?a:b, mx2=(c>d)?c:d; return ((mx1>mx2)?mx1:mx2);}
#define int ll

signed main() {
	fast;
	
	ll a[10][10];

	for(int i=0;i<10;i++) a[i][i]=0;

	ll n;
	cin>>n;

	ll aa[n];
	for(int i=0;i<n;i++) aa[i]=i;
	map<ll,ll> mp1;
	
	ll v[]={1, 2, 4, 7, 12, 20, 29, 38, 52};
	ll mul=1;
	
	for(int i=0;i<n;i++){
		ll m1;
		for(int j=0;j<n;j++){
			if(j>i){
				a[i][j]=a[j][i]=v[j-i-1]*mul;
				m1=v[j-i-1];
			}
			//assert(a[i][j]<=10000000000);
		}
		mul*=v[n-1-i-1]+v[n-1-i-2];
	}
	
	for(int i=0;i<n;i++){
		for(int j=0;j<n;j++) cout<<a[i][j]<<" ";
		cout<<endl;
	}

}