CodeForces 906C Party（状态压缩+dfs）

C. Party

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

Arseny likes to organize parties and invite people to it. However, not only friends come to his parties, but friends of his friends, friends of friends of his friends and so on. That's why some of Arseny's guests can be unknown to him. He decided to fix this issue using the following procedure.

At each step he selects one of his guests A, who pairwise introduces all of his friends to each other. After this action any two friends ofA become friends. This process is run until all pairs of guests are friends.

Arseny doesn't want to spend much time doing it, so he wants to finish this process using the minimum number of steps. Help Arseny to do it.

Input

The first line contains two integers n andm (1 ≤ n ≤ 22;) — the number of guests at the party (including Arseny) and the number of pairs of people which are friends.

Each of the next m lines contains two integersu and v (1 ≤ u, v ≤ n;u ≠ v), which means that people with numbersu and v are friends initially. It's guaranteed that each pair of friends is described not more than once and the graph of friendship is connected.

Output

In the first line print the minimum number of steps required to make all pairs of guests friends.

In the second line print the ids of guests, who are selected at each step.

If there are multiple solutions, you can output any of them.

Examples

Input

5 61 21 32 32 53 44 5

Output

22 3 

Input

4 41 21 31 43 4

Output

11 

Note

In the first test case there is no guest who is friend of all other guests, so at least two steps are required to perform the task. After second guest pairwise introduces all his friends, only pairs of guests(4, 1) and (4, 2) are not friends. Guest3 or 5 can introduce them.

In the second test case guest number 1 is a friend of all guests, so he can pairwise introduce all guests in one step.

        日常打cf……

        大致题意：给你一堆一对对的关系，然后每一个关系对代表两个人认识。然后你每次可以选择一个人i，让i认识的所有人都相互认识，即i把介绍自己所有的朋友给其他人。然后现在问你最少需要选择多少个这样的i，使得所有的人都相互认识。

        看到只有22的数据范围，很容易想到状态压缩，但是如何利用这个状态压缩却没那么好考虑。如果考虑状态压缩dp，显然无法表示连通的状态。于是考虑用搜索，22个点要么最后被选要么没有，做多2^22=400W+，在承受范围内。然后就是看如何处理这个关系了，对于每一个人，我们用一个22位的2进制数字表示他是否与其他的人有关系。对于一个人x，每次选择他的时候，枚举所有与其相连的点i，然后i的状态或上x的状态即可。这样子复杂度就是O(N*2^N)，可以通过。具体见代码：

#include<bits/stdc++.h>#define N 23using namespace std;stack<int> ans,res;int dp[N][N],n,m;void dfs(int x){    if (res.size()>=ans.size()) return;    if (x==n)    {        for(int i=0;i<n;i++)            if (dp[x][i]!=(1<<n)-1) return;        ans=res; return;    }    memcpy(dp[x+1],dp[x],sizeof(dp[x]));    dfs(x+1);    for(int i=0;i<n;i++)        if (dp[x][x]&(1<<i)) dp[x+1][i]|=dp[x][x];    res.push(x+1); dfs(x+1); res.pop();}int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)    {        ans.push(1);        dp[0][i-1]=(1<<i-1);    }    for(int i=1;i<=m;i++)    {        int x,y;        scanf("%d%d",&x,&y);        dp[0][--x]|=(1<<--y);        dp[0][y]|=(1<<x);    }    dfs(0);    printf("%d\n",ans.size());    while(!ans.empty())    {        printf("%d\n",ans.top());        ans.pop();    }    return 0;}