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;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
ll E[15][15];
int main() {
	vector<int> v(10);
	v[0] = 1;
	bitset<1000> has;
	has[1] = true;
	for (int i = 1; i < 10; ++i) {
		for (int j = v[i - 1] + 1;; ++j) {
			if (has[j]) continue;
			for (int k = 0; k < i; ++k)
				if (has[v[k] + j]) goto end;
			v[i] = j;
			break;
			end:;
		}
		has[v[i]] = true;
		for (int k = 0; k < i; ++k) has[v[k] + v[i]] = true;
	}
	ll mx = 0;
	for (int n = 2; n <= 10; ++n) {
		for (int i = 0; i + 1 < n; ++i)
			E[i][n - 1] = E[n - 1][i] = v[i] * (mx + 1);
		vector<int> perm(n);
		iota(perm.begin(), perm.end(), 0);
		do {
			ll tot = 0;
			for (int i = 0; i + 1 < n; ++i) {
				tot += E[perm[i]][perm[i + 1]];
			}
			mx = max(mx, tot);
		} while (next_permutation(perm.begin(), perm.end()));
	}
	int N;
	scanf("%d", &N);
	for (int i = 0; i < N; ++i)
		for (int j = 0; j < N; ++j)
			printf("%lld%c", E[i][j], j == N - 1 ? '\n' : ' ');
}