Codeforces Round #480 (Div. 2) E. The Number Games

E. The Number Games

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $\mathrm{nn districts\; numbered\; from 11 to nn,\; each\; district\; has\; exactly\; one\; path\; connecting\; it\; to\; every\; other\; district.\; The\; number\; of\; fans\; of\; a\; contestant\; from\; district ii is\; equal\; to 2i2i.}$

This year, the president decided to reduce the costs. He wants to remove $\mathrm{kk contestants\; from\; the\; games.\; However,\; the\; districts\; of\; the\; removed\; contestants\; will\; be\; furious\; and\; will\; not\; allow\; anyone\; to\; cross\; through\; their\; districts.}$

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 $\mathrm{nn and kk (1\le k<n\le 1061\le k<n\le 106) —\; the\; number\; of\; districts\; in\; Panel,\; and\; the\; number\; of\; contestants\; the\; president\; wishes\; to\; remove,\; respectively.}$

The next $\mathrm{n-1n-1 lines\; each\; contains\; two\; integers aa and bb (1\le a,b\le n1\le a,b\le n, a\ne ba\ne b),\; that\; describe\; a\; road\; that\; connects\; two\; different\; districts aaand bb in\; the\; nation.\; It\; is\; guaranteed\; that\; there\; is\; exactly\; one\; path\; between\; every\; two\; districts.}$

Output

Print $\mathrm{kk space-separated\; integers:\; the\; numbers\; of\; the\; districts\; of\; which\; the\; contestants\; should\; be\; removed, in\; increasing\; order\; of\; district\; number.}$

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

Note

In the first sample, the maximum possible total number of fans is ${\mathrm{22+25+26=10022+25+26=100.\; We\; can\; achieve\; it\; by\; removing\; the\; contestants\; of\; the\; districts\; 1,\; 3,\; and\; 4.}}^{}$

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 int n, k, d[1000015], vis[1000015], p[1000015];
 45 vector<int> e[1000015], e1[1000015];
 46 int f[1000015][25];
 47 
 48 void dfs(int x) {
 49     vis[x] = 1;
 50     for (int i = 0; i < e[x].size(); ++i) {
 51         int y = e[x][i];
 52         if (vis[y]) p[x] = y;
 53         else {
 54             e1[x].pb(y);
 55             dfs(y);
 56         }
 57     }
 58 }
 59 
 60 void dfs2(int x) {
 61     f[x][0] = p[x];
 62     for (int i = 1; i <= 20; ++i) {
 63         f[x][i] = f[f[x][i - 1]][i - 1];
 64     }
 65     for (int i = 0; i < e1[x].size(); ++i) {
 66         int y = e1[x][i];
 67         dfs2(y);
 68     }
 69 }
 70 vector<int> ans;
 71 bool out[1000015];
 72 void solve() {
 73     clr(vis, 0); vis[0] = 1;
 74     int now = n;
 75     while (k) {
 76         if (vis[now]) {
 77             --now;
 78             continue;
 79         }
 80         int cnt = 1;
 81         for (int h = 0, i = now; ; ) {
 82             while (!vis[f[i][h]]) {
 83                 ++h;
 84             }
 85             while (~h && vis[f[i][h]]) {
 86                 --h;
 87             }
 88             if (~h) {
 89                 cnt += 1 << h;
 90                 i = f[i][h];
 91             } else {
 92                 break;
 93             }
 94         }
 95         if (cnt <= k) {
 96             int i = now;
 97             while (!vis[i]) {
 98                 ans.pb(i);
 99                 vis[i] = 1;
100                 i = p[i];
101             }
102             k -= cnt;
103         }
104         --now;
105     }
106     for (int i = 0; i < ans.size(); ++i) {
107         out[ans[i]] = 1;
108     }
109     for (int i = 1; i <= n; ++i) {
110         if (!out[i]) printf("%d ", i);
111     }
112 }
113 int main() {
114     scanf("%d%d", &n, &k);
115     k = n - k;
116     for (int i = 1; i < n; ++i) {
117         int a, b;
118         scanf("%d%d", &a, &b);
119         e[a].pb(b), e[b].pb(a);
120     }
121     dfs(n);
122     dfs2(n);
123     solve();
124     return 0;
125 }

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