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 long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
#define sz(c) ll((c).size())
#define all(c) begin(c), end(c)
#define FOR(i,a,b) for (ll i = (a); i < (b); i++)
#define FORD(i,a,b) for (ll i = (b)-1; i >= (a); i--)
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define xx first
#define yy second
#define TR(X) ({ if(1) cerr << "TR: " << (#X) << " = " << (X) << endl; })

vl weights = {1,2,4,7,12,20,29,38,52,73};

ll max_hamilton(vvl adj) {
	ll n = sz(adj), res = 0;
	vl p(n);
	iota(all(p),0);
	do {
		ll cur = 0;
		FOR(i,1,n) cur += adj[ p[i] ][ p[i-1] ]; 
		res = max(res,cur);
	} while (next_permutation(all(p)));
	return res;
}

bool check(vvl adj) {
	ll n = sz(adj);
	vl p(n);
	iota(all(p),0);
	set<ll> s;
	do {
		ll cur = 0;
		if (p[0] > p[n-1]) continue;
		FOR(i,1,n) cur += adj[ p[i] ][ p[i-1] ];
		if (s.count(cur)) return false;
		s.insert(cur);
	} while (next_permutation(all(p)));
	return true;
}

vvl solve(ll n) {
	if (n == 2) return {{0,1},{1,0}};
	n--;
	vvl res = solve(n);
	
	ll hmax = max_hamilton(res) + (n==2);
	
	res.pb(vl(n+1));
	FOR(i,0,n) res[i].resize(n+1);
	
	FOR(i,0,n) res[i][n] = res[n][i] = weights[i] * hmax;
	return res;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll n; cin >> n;
	vvl res = solve(n);
	FOR(i,0,n) FOR(j,0,n) cout << res[i][j] << " \n"[j+1==n];
	cout << endl;
	cerr << max_hamilton(res) << endl;
	assert(check(res));
}