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;
static inline void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
static inline int ctz(unsigned x) { return __builtin_ctz(x); }
static inline int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
static inline int clz(unsigned x) { return __builtin_clz(x); }
static inline int clzll(unsigned long long x) { return __builtin_clzll(x); }
static inline int popcnt(unsigned x) { return __builtin_popcount(x); }
static inline int popcntll(unsigned long long x) {
  return __builtin_popcountll(x);
}
struct Node {
  vector<int> adj;
  int l, r;
  int d, p;
};
long long a[2 * 1000008];
Node nds[1000008];
bool take[1000008];
void add(int n, int i, int x) {
  for (; i <= n; i += i & -i) a[i] += x;
}
long long get(int i) {
  long long r = 0;
  for (; i > 0; i -= i & -i) r += a[i];
  return r;
}
void dfs(int u, int p = 0) {
  static int nxt = 0;
  nds[u].p = p;
  nds[u].l = ++nxt;
  a[nds[u].l] += nds[u].d;
  a[nds[u].l + 1] -= nds[u].d;
  for (int v : nds[u].adj) {
    if (v == p) continue;
    nds[v].d = nds[u].d + 1;
    dfs(v, u);
  }
  nds[u].r = ++nxt;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    nds[a].adj.push_back(b);
    nds[b].adj.push_back(a);
  }
  dfs(n);
  take[n] = true;
  k = n - k - 1;
  for (int i = 1; i <= 2 * n; ++i) {
    int j = i + (i & -i);
    if (j <= 2 * n) a[j] += a[i];
  }
  for (int i = n - 1; i > 0; --i) {
    if (take[i]) continue;
    long long c = get(nds[i].l);
    if (k < c) continue;
    k -= c;
    for (int u = i;;) {
      take[u] = true;
      add(2 * n, nds[u].l, -1);
      add(2 * n, nds[u].r, 1);
      u = nds[u].p;
      if (take[u]) break;
    }
    if (!k) break;
  }
  for (int i = 1; i < n; ++i)
    if (!take[i]) printf("%d ", i);
  return 0;
}
