codeforces 673D D. Bear and Two Paths(构造)

题目链接：

D. Bear and Two Paths

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bearland has n cities, numbered 1 through n. Cities are connected via bidirectional roads. Each road connects two distinct cities. No two roads connect the same pair of cities.

Bear Limak was once in a city a and he wanted to go to a city b. There was no direct connection so he decided to take a long walk, visiting each city exactly once. Formally:

• There is no road between a and b.
• There exists a sequence (path) of n distinct cities v1, v2, ..., vn that v1 = a, vn = b and there is a road between vi and vi + 1 for .

On the other day, the similar thing happened. Limak wanted to travel between a city c and a city d. There is no road between them but there exists a sequence of n distinct cities u1, u2, ..., un that u1 = c, un = d and there is a road between ui and ui + 1 for .

Also, Limak thinks that there are at most k roads in Bearland. He wonders whether he remembers everything correctly.

Given n, k and four distinct cities a, b, c, d, can you find possible paths (v1, ..., vn) and (u1, ..., un) to satisfy all the given conditions? Find any solution or print -1 if it's impossible.

Input

 

The first line of the input contains two integers n and k (4 ≤ n ≤ 1000, n - 1 ≤ k ≤ 2n - 2) — the number of cities and the maximum allowed number of roads, respectively.

The second line contains four distinct integers a, b, c and d (1 ≤ a, b, c, d ≤ n).

Output

 

Print -1 if it's impossible to satisfy all the given conditions. Otherwise, print two lines with paths descriptions. The first of these two lines should contain n distinct integers v1, v2, ..., vn where v1 = a and vn = b. The second line should contain n distinct integers u1, u2, ..., unwhere u1 = c and un = d.

Two paths generate at most 2n - 2 roads: (v1, v2), (v2, v3), ..., (vn - 1, vn), (u1, u2), (u2, u3), ..., (un - 1, un). Your answer will be considered wrong if contains more than k distinct roads or any other condition breaks. Note that (x, y) and (y, x) are the same road.

Examples

 

input

7 11
2 4 7 3

output

2 7 1 3 6 5 4
7 1 5 4 6 2 3

input

1000 999
10 20 30 40

output

-1



题意:

给出n个节点,然后给出两条路线的起点和终点,要求你构造一个无向图,使无向图中a,b之间和c,d之间均无直接相连的边,且要求这个图的边的条数不超过k;


思路：


发现n==4时怎么都不可能满足;
可以构造这样的无向图
第一条路线ac...db;

第二条路线ca...bd;
这样的边是n+1条,是最少的了;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=1e5+5;
int n,k,a,b,c,d;
int vis[1005];
int main()
{
   scanf("%d%d",&n,&k);
   scanf("%d%d%d%d",&a,&b,&c,&d);
   if(k<n+1||n==4)cout<<"-1"<<"
";
   else
   {
       vis[a]=1;
       vis[b]=1;
       vis[c]=1;
       vis[d]=1;
       printf("%d %d ",a,c);
       for(int i=1;i<=n;i++)
       {
           if(!vis[i])
           {
               printf("%d ",i);
           }
       }
       printf("%d %d
",d,b);
       printf("%d %d ",c,a);
       for(int i=1;i<=n;i++)
       {
           if(!vis[i])
           {
               printf("%d ",i);
           }
       }
       printf("%d %d 
",b,d);

   }

    return 0;
}