Problem: Let G be a simple graph. Let W be a non-empty subset of vertices. Then W is almost-k-uniform if for each pair of distinct vertices u,v ∈ W the distance between u and v is either k or k+1.

You are given a tree on n vertices. For each i between 1 and n, find the maximum size of an almost-i-uniform set.

Input

The first line contains a single integer n (2 ≤ n ≤ 5 ⋅ 10^5) – the number of vertices of the tree.

Then n-1 lines follows, the i-th of which consisting of two space separated integers u_i, v_i (1 ≤ u_i, v_i ≤ n) meaning that there is an edge between vertices u_i and v_i. 

It is guaranteed that the given graph is tree. 

Output

Output a single line containing n space separated integers a_i, where a_i is the maximum size of an almost-i-uniform set.

Examples

Input


5
1 2
1 3
1 4
4 5


Output


4 3 2 1 1


Input


6
1 2
1 3
1 4
4 5
4 6


Output


4 4 2 1 1 1

Note

Consider the first example. 

  * The only maximum almost-1-uniform set is \{1, 2, 3, 4\}. 
  * One of the maximum almost-2-uniform sets is or \{2, 3, 5\}, another one is \{2, 3, 4\}. 
  * A maximum almost-3-uniform set is any pair of vertices on distance 3. 
  * Any single vertex is an almost-k-uniform set for k ≥ 1. 



In the second sample there is an almost-2-uniform set of size 4, and that is \{2, 3, 5, 6\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int nax = 5e5 + 10;
vector<int> node[nax];
int maxdepth[nax];
int ans[nax * 2];
void dfs1(int p, int par) {
  maxdepth[p] = 0;
  for (int i : node[p]) {
    if (i == par) continue;
    dfs1(i, p);
    maxdepth[p] = max(maxdepth[p], maxdepth[i] + 1);
  }
}
int maxdeg = 0;
int diam = 0;
vector<int> dfs2(int p, int par, int topdepth = 0) {
  maxdeg = max(maxdeg, (int)node[p].size());
  diam = max(diam, topdepth);
  vector<int> pref = {0}, suf, ind, d;
  for (int i : node[p]) {
    if (i == par) continue;
    pref.push_back(maxdepth[i] + 1);
    suf.push_back(maxdepth[i] + 1);
    ind.push_back(i);
    d.push_back(maxdepth[i] + 1);
  }
  suf.push_back(0);
  for (int i = 0; i < ind.size(); i++) pref[i + 1] = max(pref[i + 1], pref[i]);
  for (int i = ind.size(); i >= 1; i--) suf[i - 1] = max(suf[i - 1], suf[i]);
  d.push_back(topdepth);
  sort(d.rbegin(), d.rend());
  vector<int> ret;
  for (int i = 1; i < d.size(); i++) {
    int v = d[i] * 2;
    ans[v] = max(ans[v], i + 1);
    ret.push_back(d[i]);
    v = d[i] * 2 + 1 - (d[i - 1] == d[i]) * 2;
    ans[v] = max(ans[v], i + 1);
  }
  vector<int> comb;
  for (int i = 0; i < ind.size(); i++) {
    vector<int> c =
        dfs2(ind[i], p, max(topdepth + 1, max(pref[i], suf[i + 1]) + 1));
    for (int j = 0; j < c.size(); j++) {
      if (j == comb.size())
        comb.push_back(c[j]);
      else
        comb[j] = max(comb[j], c[j]);
    }
  }
  {
    int a = 2, b = 2;
    while (a < d.size() && b < comb.size()) {
      int v = min(d[a], comb[b]);
      ans[v * 2] = max(ans[v * 2], a + b + 2 - 2);
      if (b + 1 == comb.size() || a + 1 < d.size() && d[a + 1] >= comb[b + 1]) {
        a++;
      } else {
        b++;
      }
    }
  }
  return d;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    node[a].push_back(b);
    node[b].push_back(a);
  }
  fill_n(ans + 1, n, 1);
  dfs1(0, -1);
  dfs2(0, -1);
  ans[1] = maxdeg + 1;
  for (int i = n - 2; i >= 1; i--) {
    ans[i] = max(ans[i], ans[i + 2]);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
}