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 pair<int, int> pii;
typedef long long ll;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()
#define dump(x) cout << #x << " = " << (x) << endl
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

template<class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}
 
template<class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
	os<<"{";
	rep(i, v.size()) {
		if (i) os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

const int maxn = 10;

ll d[maxn][maxn];

int main() {
	vector<ll> dp(maxn);
	dp[0] = 1;
	dp[1] = 2;

	for (int i = 2; i < maxn; ++i) {
		set<ll> st;
		rep(j, i) rep(k, j) {
			st.insert(dp[k] + dp[j]);
		}

		ll c = dp[i-1] + 1;
		while (true) {
			bool ok = true;
			rep(j, i) {
				if (st.count(c + dp[j]) || st.count(c)) {
					ok = false; break;
				}
			}
			if (ok) {
				break;
			}
			c++;
		}
		dp[i] = c;
	}

	//cout << dp << endl;

	int N; cin >> N;

	d[0][1] = d[1][0] = 1;

	for (int i = 2; i < N; ++i) {
		ll c = 0;
		vi pm;
		rep(j, i) pm.pb(j);
		vector<ll> cst;

		do {
			ll s = 0;
			rep(j, (int)pm.size()-1) {
				s += d[pm[j]][pm[j+1]];
			}
			cst.pb(s);
			c = max(c, s);
		} while (next_permutation(ALL(pm)));
		++c;
		//dump(cst);
		//sort(ALL(cst));
		//cst.erase(unique(ALL(cst)), cst.end());
		//dump(cst.size());
		//cerr << i << " " << c << endl;

		rep(j, i) d[j][i] = d[i][j] = c * dp[j];
	}

	rep(i, N) {
		rep(j, N) {
			printf("%lld%c", d[i][j], j == N-1 ? '\n' : ' ');
		}
	}

	return 0;
}