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>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) (((x) < 0)?-(x):(x))
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

using cat = long long;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

cat W[10][10];

void upd(int a, int b, cat hint = 1) {
	int P[10];
	for(int i = 0; i < 10; i++) P[i] = i;
	vector<cat> sum_c, sum_l;
	while(true) {
		if(P[0] < P[9]) {
			cat s = 0;
			int l = 0;
			for(int i = 0; i < 9; i++) {
				if(P[i] == a && P[i+1] == b) l++;
				else if(P[i] == b && P[i+1] == a) l++;
				else s += W[P[i]][P[i+1]];
			}
			if(l == 0) sum_c.push_back(s);
			else sum_l.push_back(s);
		}
		if(!next_permutation(P, P+10)) break;
	}
	sort(begin(sum_c), end(sum_c));
	sort(begin(sum_l), end(sum_l));
	vector<int> pos_c(sum_l.size(), 0);
	priority_queue< pair<cat, int>, vector< pair<cat, int> >, greater< pair<cat, int> > > q;
	for(int i = 0; i < (int)sum_l.size(); i++) {
		if(i > 0) pos_c[i] = pos_c[i-1];
		while(pos_c[i] < (int)sum_c.size() && sum_c[pos_c[i]] < sum_l[i]+hint) pos_c[i]++;
		if(pos_c[i] < (int)sum_c.size()) q.push({sum_c[pos_c[i]]-sum_l[i], i});
	}
	cat x = hint;
	while(!q.empty()) {
		pair<cat, int> p = q.top();
		if(x < p.ff) break;
		q.pop();
		while(true) {
			pos_c[p.ss]++;
			q.push({sum_c[pos_c[p.ss]]-sum_l[p.ss], p.ss});
			p = q.top();
			if(p.ff > x) break;
			q.pop();
		}
		x++;
	}
	if(q.empty()) return;
	cerr << x << " " << max(sum_c.back(), sum_l.back()+x)*1e-11 << endl;
	W[a][b] = W[b][a] = x;
}

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N;
	cin >> N;
	memset(W, 0, sizeof(cat)*100);
	int seq[11] = {1, 2};
	for(int i = 2; i <= 10; i++) seq[i] = seq[i-1]+seq[i-2]+1;
	cat prod = 1;
	int n = N-1;
	while(n > 0) {
		for(int i = 0; i < n; i++) W[i][n] = W[n][i] = prod * seq[i];
		prod *= seq[n];
		n--;
	}
	if(N == 10) {
		W[1][2] = W[2][1] = 3;         //upd(1, 2);
		W[0][1] = W[1][0] = 214;       //upd(0, 1);
		W[0][2] = W[2][0] = 95963;     //upd(0, 2, 50000);
		W[2][3] = W[3][2] = 1002019965;//upd(2, 3, 1002000000);
		W[1][3] = W[3][1] = 2004429048;//upd(1, 3, 2004000000);
		W[0][3] = W[3][0] = 4008331713;//upd(0, 3, 4008000000);
	}
	for(int i = 0; i < N; i++)
		for(int j = 0; j < N; j++) cout << W[i][j] << ((j+1 == N) ? "\n" : " ");
	return 0;
}

// look at my code
// my code is amazing