Codeforces 673D Bear and Two Paths【贪心】

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本文介绍了一个有趣的问题，如何在限定条件下构建一个包含特定路径的无向图。文章给出了实现思路和AC代码，确保两个指定点间存在一条遍历所有节点的路径，并且满足边数限制。

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, ..., v_n that v₁ = a, v_n = b and there is a road between v_i and v_i + 1 for $\text{[math]}$ .

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, ..., u_n that u₁ = c, u_n = d and there is a road between u_i and u_i + 1 for $\text{[math]}$ .

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, ..., v_n) and (u₁, ..., u_n) 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, ..., v_n where v₁ = a and v_n = b. The second line should contain n distinct integers u₁, u₂, ..., u_n where u₁ = c and u_n = d.

Two paths generate at most 2n - 2 roads: (v₁, v₂), (v₂, v₃), ..., (v_n - 1, v_n), (u₁, u₂), (u₂, u₃), ..., (u_n - 1, u_n). 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.

$\text{[math]}$

题目大意：

给你一个包含N个点的无向图，要求我们用小于等于K条边来构造出这个图。

这个图有两对起点和终点：a，b，c，d；

现在要求构造出的图，a和b之间没有直接相连的边，c和d之间没有直接相连的边。

但是从a到b有一条路径，a作为起点，b作为终点，路径上包含所有点（1.2.3.4.5..............N）。

同理，从c到d也有一条路径，c作为起点，d作为终点，路径上也包含所有点。

思路：

1、因为边数有限制，我们肯定是想用最少的边数来完成这个任务。

对于从a到b的路径，用最少的边肯定就是让a作为起点，b作为终点的同时，其他点作为中间点，构成一个欧拉路径。

比如样例我们就可以很简单的构造出来一种可行答案：2 1 3 5 6 74

2、但是现在又要求要有一条从c到d的路径。那么我们此时希望从c到d的这条欧拉路经和从a到b的路径分享更多的重边。

通过简单的设想，我们不难发现，当我们设定从a到b的路径为：

a c u............v d b

同时设定从c到d的路径为：

c a u............v b d

的时候，我们只要在从a到b的欧拉路径的基础上，增加两条边就能够使得整个图满足题目要求的条件【（a，u），（v，b）】。

3、注意N==4的时候，无论怎样设定都会有相邻的情况。

Ac代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int vis[100060];
int ans[100060];
int ans2[100060];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(vis,0,sizeof(vis));
        int a,b,c,d;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        if(n==4||m<n+1)
        {
            printf("-1\n");
            continue;
        }
        ans[1]=a;
        ans[n]=b;
        ans[2]=c;
        ans[n-1]=d;
        int cnt=3;
        for(int i=1;i<=n;i++)
        {
            if(i!=a&&i!=b&&i!=c&&i!=d)ans[cnt++]=i;
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
        ans2[1]=c;
        ans2[n]=d;
        ans2[2]=a;
        ans2[n-1]=b;
        cnt=3;
        for(int i=1;i<=n;i++)
        {
            if(ans[i]!=a&&ans[i]!=b&&ans[i]!=c&&ans[i]!=d)ans2[cnt++]=ans[i];
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d ",ans2[i]);
        }
        printf("\n");
    }
}