Codeforces 673D Bear and Two Paths【贪心】

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 cityb. There was no direct connection so he decided to take a long walk, visiting each cityexactly once. Formally:

• There is no road between a and b.
• There exists a sequence (path) of n distinct citiesv₁, v₂, ..., v_n thatv₁ = a,v_n = b and there is a road betweenv_i andv_i + 1 for.

On the other day, the similar thing happened. Limak wanted to travel between a cityc and a city d. There is no road between them but there exists a sequence ofn distinct cities u₁, u₂, ..., u_n thatu₁ = c,u_n = d and there is a road betweenu_i andu_i + 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 citiesa, 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 andk (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 integersa, 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 containn distinct integers v₁, v₂, ..., v_n wherev₁ = a andv_n = b. The second line should containn distinct integers u₁, u₂, ..., u_n whereu₁ = c andu_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

7 112 4 7 3

Output

2 7 1 3 6 5 47 1 5 4 6 2 3

Input

1000 99910 20 30 40

Output

-1

Note

In the first sample test, there should be 7 cities and at most11 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");    }}