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;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 5;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power(((((a) % mod) * ((a) % mod)) % mod), b / 2) % mod;
  else
    return (((a) % mod) *
            (power(((((a) % mod) * ((a) % mod)) % mod), b / 2) % mod)) %
           mod;
}
inline int64_t inverse(int64_t a, int64_t md) {
  a %= md;
  if (a < 0) a += md;
  int64_t b = md, u = 0, v = 1;
  while (a) {
    int64_t t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
const int ce = 1e9 + 7;
int64_t fast_mod(int64_t input) { return input < ce ? input : input % ce; }
vector<int> eu;
int he[2000001];
int par[2000001];
vector<int> v[2000001];
bool vis[2000001];
vector<int> fin;
void dfs(int u, int pari) {
  eu.push_back(u);
  for (auto i : v[u]) {
    if (i == pari) continue;
    he[i] = he[u] + 1;
    par[i] = u;
    dfs(i, u);
  }
  eu.push_back(-u);
}
int st[2000001];
int en[2000001];
int tot, rem;
void dfs1(int u, int pari) {
  if (vis[u]) return;
  fin.push_back(u);
  vis[u] = 1;
  tot--;
  rem--;
  for (auto i : v[u]) {
    if (i == pari) continue;
    dfs1(i, u);
  }
}
int tree[8000011];
void update(int node, int start, int end, int idx, int val) {
  if (start == end) {
    tree[node] = val;
  } else {
    int mid = (start + end) / 2;
    if (start <= idx and idx <= mid) {
      update(2 * node, start, mid, idx, val);
    } else {
      update(2 * node + 1, mid + 1, end, idx, val);
    }
    tree[node] = tree[2 * node] + tree[2 * node + 1];
  }
}
int query(int node, int start, int end, int l, int r) {
  if (r < start or end < l) {
    return 0;
  }
  if (l <= start and end <= r) {
    return tree[node];
  }
  int mid = (start + end) / 2;
  int p1 = query(2 * node, start, mid, l, r);
  int p2 = query(2 * node + 1, mid + 1, end, l, r);
  return (p1 + p2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  par[n] = 0;
  he[n] = 1;
  memset(tree, 0, sizeof(tree));
  memset(vis, false, sizeof(vis));
  dfs(n, 0);
  for (int i = 0; i < eu.size(); i++) {
    if (eu[i] > 0)
      st[eu[i]] = i + 1;
    else
      en[abs(eu[i])] = i + 1;
  }
  int sum = 0;
  tot = n, rem = k;
  for (int i = n; i >= 1; --i) {
    if (vis[i]) continue;
    int pr = query(1, 1, 2 * n, 1, st[i]);
    int ne = sum - pr + he[i];
    if (tot - ne >= rem) {
      sum = ne;
      int x = i;
      while (x) {
        if (vis[x]) break;
        update(1, 1, 2 * n, st[x], 1);
        update(1, 1, 2 * n, en[x], -1);
        vis[x] = 1;
        x = par[x];
      }
    } else {
      dfs1(i, par[i]);
    }
  }
  sort(fin.begin(), fin.end());
  for (auto i : fin) cout << i << " ";
}
