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:
// Solution #1 : Max path length = 83 907 014 282
 
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
 
long long N, a[11][11], r = 1, e[10] = { 1, 2, 4, 7, 12, 20, 29, 38, 52 };
 
long long solve() {
	vector<long long>E, F;
	for (int i = 0; i < N; i++) E.push_back(i);
 
	do {
		long long s = 0;
		for (int i = 0; i < N - 1; i++) {
			s += a[E[i]][E[i + 1]];
		}
		F.push_back(s);
	} while (next_permutation(E.begin(), E.end()));
 
	sort(F.begin(), F.end());
	F.erase(unique(F.begin(), F.end()), F.end());
 
	long long LIM = 1;
	for (int i = 1; i <= N; i++) LIM *= i;
	LIM /= 2;
 
	//cout << F.size() << " " << LIM << endl;
	if (F.size() == LIM) return F[F.size() - 1];
	return (1LL << 60);
}
 
int main() {
	cin >> N; a[0][1] = 1; a[1][0] = 1; a[0][2] = 2; a[2][0] = 2; a[1][2] = 3; a[2][1] = 3;
 
	for (int i = 3; i < N; i++) {
		long long NN = N; N = i;
		long long V = solve(); N = NN;
		r = V + 1;
		for (int j = 0; j < i; j++) {
			long long cost = r * e[j];
			a[i][j] = cost; a[j][i] = cost;
		}
	}
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) { if (j) cout << " "; cout << a[i][j]; } cout << endl;
	}
	return 0;
}