Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D

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

Note

In the first sample test, there should be 7 cities and at most 11 roads. The provided sample solution generates 10 roads, as in the drawing. You can also see a simple path of length n between 2 and 4, and a path between 7 and 3.

题意： 给你n个点  最多 k条边 组成一个图 使得a~b  c~d （无直接边相连）  存在一条经过每一点的路径

输出路径  若无法满足 输出-1

题解： 当n==4||n+1>k 时 无法满足 输出-1

先 取a,b,c,d 以及任意一点 共5个点  按题目要求摆放并画出

其余n-5个点随意添加  为了方便  全部加在了现有的一条边上 然后输出

 #include<bits/stdc++.h>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<map>
 #define ll __int64
 #define pi acos(-1.0)
 using namespace std;
 int n,k;
 int a,b,c,d;
 int exm;
 map<int,int>mp;
 int main()
 {
     mp.clear();
     scanf("%d %d",&n,&k);
     if(n==)
     {
         cout<<"-1"<<endl;
         return ;
     }
     scanf("%d %d %d %d",&a,&b,&c,&d);
     mp[a]=;
     mp[b]=;
     mp[c]=;
     mp[d]=;
     for(int i=;i<=n;i++)
     {
         if(mp[i]==)
         {
         exm=i;
         mp[i]=;
         break;
         }
     }
     if(+(n-)>k)
     {
         cout<<"-1"<<endl;
        return ;
     }
     cout<<a;
     for(int i=;i<=n;i++)
      {
          if(mp[i]==)
          cout<<" "<<i;
      }
      cout<<" "<<d<<" "<<exm<<" "<<c<<" "<<b<<endl;
      cout<<c<<" "<<b<<" "<<exm<<" "<<a;
     for(int i=;i<=n;i++)
      {
          if(mp[i]==)
          cout<<" "<<i;
      }
      cout<<" "<<d<<endl;
     return ;
 }
 close