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:
/* cerberus97 - Hanit Banga */

#include <iostream>
#include <iomanip>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <chrono>
#include <random>

using namespace std;

#define pb push_back
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL)

typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;

const int N = 15;
const ll MX = (1e11 / 9);

ll e[N][N];
set<ll> st[N];
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

void solve(int n);
ll get_max(int n);
bool check(int n);

int main() {
	fast_cin();
	int n; cin >> n;
	solve(n);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= n; ++j) {
			cout << e[i][j] << ' ';
		}
		cout << '\n';
	}
}

void solve(int n) {
	vector<int> sum_free = {0, 1, 2, 4, 7, 12, 20, 29, 38, 52, 73};
	e[1][2] = e[2][1] = 1;
	ll M = 1;
	for (int i = 3; i <= n; ++i) {
		for (int j = 1; j < i; ++j) {
			e[i][j] = e[j][i] = M * sum_free[(i == 3 ? j + 1 : j)];
		}
		M = get_max(i);
	}
	// cout << M / (1e11) << endl;
	// cout << M << endl;
	// check(n);
	// cout << check(n) << endl;
}

ll get_max(int n) {
	vector<int> perm;
	for (int i = 1; i <= n; ++i) {
		perm.pb(i);
	}
	ll ans = 0;
	do {
		if (perm[0] > perm[n - 1]) {
			continue;
		}
		ll sum = 0;
		for (int i = 0; i < n - 1; ++i) {
			sum += e[perm[i]][perm[i + 1]];
		}
		ans = max(ans, sum);
	} while (next_permutation(perm.begin(), perm.end()));
	return ans;
}

bool check(int n) {
	vector<int> perm;
	for (int i = 1; i <= n; ++i) {
		perm.pb(i);
	}
	set<ll> all_sums;
	int ctr = 0;
	do {
		if (perm[0] > perm[n - 1]) {
			continue;
		}
		++ctr;
		ll sum = 0;
		for (int i = 0; i < n - 1; ++i) {
			sum += e[perm[i]][perm[i + 1]];
		}
		all_sums.insert(sum);
	} while (next_permutation(perm.begin(), perm.end()));
	cout << ctr << ' ' << all_sums.size() << endl;
	return (ctr == all_sums.size());
}