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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <stdio.h>
#include <time.h>
#include <numeric>
#include <random>
#include <unordered_map>
#include <unordered_set>
#define all(a) a.begin(),a.end()
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define pb push_back
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> P;
typedef complex<ld> com;
constexpr int inf = 1000000010;
constexpr ll INF = 1000000000000000010;
constexpr ld eps = 1e-12;
constexpr ld pi = 3.141592653589793238;
template<class T, class U> inline bool chmax(T &a, const U &b) { if (a < b) { a = b; return true; } return false; }
template<class T, class U> inline bool chmin(T &a, const U &b) { if (a > b) { a = b; return true; } return false; }

struct str {
	ll dis, x, y;
};

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);

	int n;
	cin >> n;
	vector<vector<ll>> ans(10, vector<ll>(10));
	ans[0][1] = ans[1][0] = 1;
	ans[0][2] = ans[2][0] = 2;
	ans[1][2] = ans[2][1] = 3;
	vector<str> vec;
	vec.pb({ 1,0,1 }); vec.pb({ 2,0,2 }); vec.pb({ 3,1,2 });
	for (int i = 3; i < 10; i++) {
		for (int j = 0; j < i; j++) {
			ll sum = 1;
			vector<int> check(10);
			int cnt = i;
			if (j == 0) cnt--;
			int small = cnt - 1; int big = cnt;
			for (int s = 0; s < vec.size(); s++) {
				if (small == 0) break;
				sum -= vec[s].dis;
				check[vec[s].x]++; check[vec[s].y]++;
				small--;
			}
			rep(c, 10) check[c] = 0;
			for (int s = vec.size() - 1; s >= 0; s--) {
				if (big == 0) break;
				if (check[vec[s].x] < 2 && check[vec[s].y] < 2) {
					sum += vec[s].dis;
					check[vec[s].x]++; check[vec[s].y]++;
					big--;
				}
			}
			vec.pb({ sum,i,j });
			ans[i][j] = ans[j][i] = sum;
		}
	}
	rep(i, n) {
		rep(j, n) {
			cout << ans[i][j] << ' ';
		}
		cout << '\n';
	}
}