Codeforces 323 B Tournament-graph

Discription

In this problem you have to build tournament graph, consisting of n vertices, such, that for any oriented pair of vertices (v, u) (v ≠ u) there exists a path from vertexv to vertex u consisting of no more then two edges.

A directed graph without self-loops is a tournament, if there is exactly one edge between any two distinct vertices (in one out of two possible directions).

Input

The first line contains an integer n (3 ≤ n ≤ 1000), the number of the graph's vertices.

Output

Print -1 if there is no graph, satisfying the described conditions.

Otherwise, print n lines with n integers in each. The numbers should be separated with spaces. That is adjacency matrix a of the found tournament. Consider the graph vertices to be numbered with integers from 1 to n. Then a_v, u = 0, if there is no edge from v to u, and a_v, u = 1 if there is one.

As the output graph has to be a tournament, following equalities must be satisfied:

• a_v, u + a_u, v = 1 for each v, u (1 ≤ v, u ≤ n; v ≠ u);
• a_v, v = 0 for each v (1 ≤ v ≤ n).

Example

Input

3

Output

0 1 0
0 0 1
1 0 0

Input

4

Output

-1


构造题，比较迷。
首先我们如果有了n个点的合法图是很容易构造出n+2个点的合法图的。
首先可以把1的入点看成一类点，1的出点看成一类点，然后再加上1本身和n+1和n+2，我们现在就有了5个点。
而我们的目的是让任意一对点都在至少一个三元环 (形如a->b,b->c,c->a) 出现。
所以直接xjb构造就行了，这个其实可以不用手算，直接写个程序跑一跑也是可以的2333(我是不会告诉你们我的连边方案就是电脑枚举出来的哈哈)

但是发现4没有答案，导致我一开始以为偶数都是gg的然后就WA了。。。
后来写了个搜索发现6是有答案的2333，所以是偶数的话特判完了之后直接从6的图往后跑就行了，n=6的图可以搜索也可以手玩(反正写搜索就当练暴力了2333)

#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
using namespace std;
int a[maxn][maxn];
int n,m,k,tone[maxn];
int main(){
	scanf("%d",&n);
	if(!(n&1)){
		if(n<6){
			puts("-1");
			return 0;
		}
		
		a[1][5]=a[1][6]=1;
		a[2][1]=a[2][6]=1;
		a[3][1]=a[3][2]=a[3][5]=1;
		a[4][1]=a[4][2]=a[4][3]=1;
		a[5][2]=a[5][4]=1;
		a[6][3]=a[6][4]=a[6][5]=1;
		
		tone[2]=tone[3]=tone[4]=1;
		
		for(int i=7;i<=n;i+=2){
			tone[i]=1;
			a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1;
			for(int j=2;j<i;j++){
				if(tone[j]) a[j][i]=1,a[i+1][j]=1;
				else a[j][i+1]=1,a[i][j]=1;
			}
		}
	}
	else{
		a[1][2]=a[2][3]=a[3][1]=1;
		tone[3]=1;
		for(int i=4;i<=n;i+=2){
			tone[i]=1;
			a[i][1]=1,a[1][i+1]=1,a[i+1][i]=1;
			for(int j=2;j<i;j++){
				if(tone[j]) a[j][i]=1,a[i+1][j]=1;
				else a[j][i+1]=1,a[i][j]=1;
			}
		}
	}	

	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++) printf("%d ",a[i][j]);
		puts("");
	}

	
	/*
	for(int i=1;i<=n;i++)
	    for(int j=1;j<=n;j++) a[i][j]=(a[i][j]?1:1<<30);
	for(int i=1;i<=n;i++) a[i][i]=0;
	
	for(int k=1;k<=n;k++)
	    for(int i=1;i<=n;i++)
	        for(int j=1;j<=n;j++) if(a[i][k]+a[k][j]<a[i][j]) a[i][j]=a[i][k]+a[k][j];
	        
	for(int i=1;i<=n;i++)
	    for(int j=1;j<=n;j++) if(a[i][j]>2) puts("No");
	*/
	
	return 0;
}