E. The Number Games

The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant.

The nation has $n$

districts numbered from $1$ to $n$, each district has exactly one path connecting it to every other district. The number of fans of a contestant from district $i$ is equal to $2^i$

.

This year, the president decided to reduce the costs. He wants to remove $k$

contestants from the games. However, the districts of the removed contestants will be furious and will not allow anyone to cross through their districts.

The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants.

Which contestants should the president remove?

Input

The first line of input contains two integers $n$

and $k$ ( $1\le k<n\le {10}^6$

) — the number of districts in Panel, and the number of contestants the president wishes to remove, respectively.

The next $n-1$

lines each contains two integers $a$ and $b$ ( $1\le a,b\le n$, $a\ne b$), that describe a road that connects two different districts $a$ and $b$

in the nation. It is guaranteed that there is exactly one path between every two districts.

Output

Print $k$

space-separated integers: the numbers of the districts of which the contestants should be removed, in increasing order of district number.

Examples

Input

Copy

6 3
2 1
2 6
4 2
5 6
2 3

Output

Copy

1 3 4

Input

Copy

8 4
2 6
2 7
7 8
1 2
3 1
2 4
7 5

Output

Copy

1 3 4 5

#include <bits/stdc++.h>
using namespace std;

const int maxn=1e6+10;
vector<int> vec[maxn];
/*
这道题我有思路，但是不会写
因为每个数字各不相同，所以如果能选择大的，绝不会选择小的，因为下面的全都加起来，也不如一个
高位，所以就是怎样来选择的问题了，nlogn的复杂度，从大到小遍历，如果可以选，我们一定选
并且把路径上的点，全都选上，这样才可以，所以用最近公共祖先来做，其实就是记忆化，每次直接查找就可以了
*/

int vis[maxn],in[maxn];

int p[maxn][21];

void dfs(int o,int fa){
    p[o][0]=fa;vis[o]=1;
    for(int i=1;i<=20&&p[p[o][i-1]][i-1];i++)p[o][i]=p[p[o][i-1]][i-1];
    int sz=vec[o].size();
    for(int i=0;i<sz;i++){
        int v=vec[o][i];
        if(vis[v])continue;
        dfs(v,o);
    }
}

int cal(int o,int k){
    int ans=o;
    for(int i=20;k&&i>=0;i--){
        if(k>=(1<<i)){
            k-=(1<<i);
            ans=p[ans][i];
        }
    }
    return ans;
}
int ans[maxn];

int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    k=n-k;
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d %d",&u,&v);
        vec[u].push_back(v),vec[v].push_back(u);
    }
    dfs(n,n);
/*
    for(int i=1;i<=6;i++){
        for(int j=0;j<=2;j++){
            printf(" i:%d j:%d %d",i,j,p[i][j]);
        }
        printf("\n");
    }
*/
    in[n]=1;k--;
    for(int i=n;i>=1&&k;i--){
        if(in[i])continue;
        int rt=cal(i,k);
        //printf("i:%d rt:%d k:%d\n",i,rt,k);
        if(in[rt]){
            int fa=i;
            while(!in[fa]){
                in[fa]=1;
                fa=p[fa][0];
                k--;
            }
        }
    }
    int cnt=0;
    for(int i=1;i<=n;i++){
        if(!in[i])ans[cnt++]=i;
    }
    for(int i=0;i<cnt;i++){
        if(!i)printf("%d",ans[i]);
        else printf(" %d",ans[i]);
    }
    printf("\n");
    return 0;
}