D. Bear and Two Paths(贪心构造)

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, ..., v_n that v₁ = a, v_n = b and there is a road between v_i and v_i + 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, ..., u_n that u₁ = c, u_n = d and there is a road between u_i and u_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 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.

题意：看a,b之间，c,d之间是否遍历所有顶点且一次后有路，且路的数目不超过K.

题解：

构造一条a,c……d,b的路

和一条   c,a……b,d的路，最少需要n+1条，所以k>=n+1  且n == 4时，a,b之间和c,d之间都不能有通路，所以n == 4怎么都不行。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int vis[1005];
void solve(){
    int n,k;
    int a,b,c,d;
    scanf("%d %d",&n,&k);
    scanf("%d%d%d%d",&a,&b,&c,&d);
    if(n == 4 || k<n+1) printf("-1
");
    else{
        vis[a] = vis[b] = vis[c] = 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);
    }
}
int main()
{
    solve();
    return 0;
}