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;
#define DBG(v) cerr << #v << " = " << (v) << endl;

vector<int> f() {
	vector<int> v;
	set<int> s;

	auto ok = [&](const int i) {
		if(s.count(i))
		return false;
		for(int x : v)
		if(s.count(x +i))
		return false;
		return true;
	};

	for (int i = 1; (int) v.size() < 10; ++i)
		if (ok(i)) {
			s.insert(i);
			for (int x : v)
				s.insert(x + i);
			v.push_back(i);
		}

	assert(s.size() == 10 + 10 * 9 / 2);
	return v;
}

ll M[111][111];

ll getMx(const ll n) {
	vector<int> v(n);
	for (int i = 0; i < n; ++i)
		v[i] = i + 1;
	ll mx = 0;
	do {
		ll temp = 0;
		for (int i = 1; i < n; ++i)
			temp += M[v[i - 1]][v[i]];
		mx = max(mx, temp);
	} while (next_permutation(v.begin(), v.end()));
	return mx;
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);

	int n;
	cin >> n;
	assert(2 <= n && n <= 10);
	M[1][2] = M[2][1]= 1;

	const vector<int> v = f();

	for (int i = 3; i <= n; ++i) {
		ll mx = getMx(i - 1);

		for (int j = 1; j < i; ++j)
			M[i][j] = M[j][i] =(mx + 1) * v[j - 1];
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j)
			cout << M[i][j] << ' ';
		cout << endl;
	}
	cerr << "mx: " << getMx(n);
	return 0;
}