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 ≤ k < n ≤ 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 ≤ a, b ≤ n, a ≠ 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

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


Output

1 3 4


Input

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


Output

1 3 4 5

Note

In the first sample, the maximum possible total number of fans is 2^2 + 2^5 + 2^6 = 100. We can achieve it by removing the contestants of the districts 1, 3, and 4.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 100;
long long tree[N];
bool a[N];
int sz[N];
int mapping[N];
int par[N];
long long n;
void print() {
  for (int i = 1; i <= n; i++) {
    if (a[i] == false) printf("%d ", i);
  }
}
long long query(int idx) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= idx & (-idx);
  }
  return sum;
}
void print(int n) {
  cout << "--------------------\n";
  for (int i = 1; i <= n; i++) {
    cout << i << " " << query(mapping[i]) << " " << sz[i] << " mapp "
         << mapping[i] << endl;
    ;
  }
  cout << "-----------------\n";
}
void update(long long idx, long long x, long long n) {
  while (idx <= n) {
    tree[idx] += x;
    idx += idx & (-idx);
  }
}
void rangeUpdate(long long x, long long y, long long val, long long n) {
  update(x, val, n);
  update(y + 1, -val, n);
}
int cc = 1;
int len;
vector<int> v[N];
void dfs(int x, int pr, int dis) {
  len++;
  par[x] = pr;
  mapping[x] = cc;
  rangeUpdate(cc, cc, dis, n);
  cc++;
  int st = len;
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (y != pr) dfs(y, x, dis + 1);
  }
  sz[x] = len - st;
}
int main() {
  long long q, i, j = 0, temp, t, k, ans = 0, sum = 0, x, y, z, cnt = 0, m,
                  fg = 0, mx = 0, mx1 = 0, mn = 8000000000000000000,
                  mn1 = 8000000000000000000;
  scanf("%lld %lld", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    scanf("%lld %lld", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(n, n, 0);
  if (k == 0) {
    print();
    return 0;
  }
  a[n] = true;
  k = n - k;
  k--;
  for (int i = n - 1; i >= 1; i--) {
    if (!k) break;
    if (a[i]) continue;
    int val = query(mapping[i]);
    if (val > k) continue;
    k -= val;
    j = i;
    while (!a[j]) {
      a[j] = true;
      rangeUpdate(mapping[j], mapping[j] + sz[j], -1, n);
      j = par[j];
    }
  }
  print();
}
