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>
#define mod 1000000007
#define mod998 998244353
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

const ll x[10] = { 1,2,4,7,12,20,33,54,88,143 };

int N;
ll a[10][10], b;

int main() {
	cin >> N;
	b = 1;
	for (int i = 0; i < N; ++i) {
		for (int j = i + 1; j < N; ++j) {
			a[i][j] = a[j][i] = x[j - i - 1] * b + 1;
		}
		b *= x[9 - i] - 1;
	}
	a[9][8] = a[8][9] = 1;
	for (int i = 0; i < N; ++i) {
		for (int j = 0; j < N; ++j) {
			cout << a[i][j];
			if (j != N - 1)cout << sp;
		}
		cout << endl;
	}
	/*
	vector<int>v{ 0,1,2,3,4,5,6,7,8,9 };
	map<ll, vector<int>>mp;
	do {
		if (v.front() > v.back())continue;
		ll t = 0;
		for (int i = 0; i < 9; ++i) {
			t += a[v[i]][v[i + 1]];
		}
		if (t > 1e11) {
			cout << t << endl;
			for (int i : v) {
				cout << i << sp;
			}
			cout << endl;
		}
		if (mp.find(t) != mp.end()) {
			cout << t << endl;
			for (int i : mp[t]) {
				cout << i << sp;
			}
			cout << endl;
			for (int i : v) {
				cout << i << sp;
			}
			cout << endl;
		}
		else {
			mp[t] = v;
		}
	} while (next_permutation(v.begin(), v.end()));*/
}