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:
// Copyright: lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<long long,long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i,j,k)  for(register int i=(int)(j);i<=(int)(k);i++)
#define rrep(i,j,k) for(register int i=(int)(j);i>=(int)(k);i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)

ll read(){
	ll x=0,f=1;char c=getchar();
	while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
	while(c>='0' && c<='9'){x=x*10+c-'0';c=getchar();}
	return x*f;
}

int a[] = {0, 1, 2, 4, 7, 12, 20, 29, 38, 52, 73};
ll w[15][15];

void work() {
	int n = read();
	w[1][2] = w[2][1] = 1;
	rep(i, 3, n) {
		ll s = 0;
		int p[15];
		rep(x, 1, i - 1) p[x] = x;
		do {
			ll nw = 0;
			rep(x, 1, i - 2) nw += w[p[x]][p[x + 1]];
			s = max(s, nw);
		} while (next_permutation(p + 1, p + i));
		s++;
		rep(x, 1, i - 1) w[x][i] = w[i][x] = s * a[x];
	}
	rep(i, 1, n) {
		rep(j, 1, n) printf("%lld ", w[i][j]);
		puts("");
	}
}

int main(){
	#ifdef LZT
		freopen("in","r",stdin);
	#endif

	work();

	#ifdef LZT
		Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
	#endif
}