A tree is an undirected graph with exactly one simple path between each pair of vertices. We call a set of simple paths k-valid if each vertex of the tree belongs to no more than one of these paths (including endpoints) and each path consists of exactly k vertices.

You are given a tree with n vertices. For each k from 1 to n inclusive find what is the maximum possible size of a k-valid set of simple paths.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vertices in the tree.

Then following n - 1 lines describe the tree, each of them contains two integers v, u (1 ≤ v, u ≤ n) — endpoints of the corresponding edge.

It is guaranteed, that the given graph is a tree. 

Output

Output n numbers, the i-th of which is the maximum possible number of paths in an i-valid set of paths.

Examples

Input

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


Output

7
3
2
1
1
1
1



Input

6
1 2
2 3
2 4
1 5
5 6


Output

6
2
2
1
1
0

Note

One way to achieve the optimal number of paths for the second sample is illustrated in the following picture:

<image>
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 30;
const int MAXL = 320;
vector<int> g[MAXN];
pair<int, int> dfs(int u, int p, const int &k) {
  pair<int, int> ret;
  int mx1 = 0, mx2 = 0;
  for (int v : g[u]) {
    if (v == p) continue;
    auto tmp = dfs(v, u, k);
    ret.first += tmp.first;
    if (tmp.second >= mx1) {
      mx2 = mx1;
      mx1 = tmp.second;
    } else if (tmp.second > mx2)
      mx2 = tmp.second;
  }
  if (mx1 + mx2 + 1 >= k) {
    ret.first++;
    ret.second = 0;
  } else
    ret.second = mx1 + 1;
  return ret;
}
vector<int> ans(MAXN, -1);
inline int get(const int &k) {
  return (ans[k] == -1 ? ans[k] = dfs(1, -1, k).first : ans[k]);
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= min(n, MAXL); i++) get(i);
  int prv = n + 1;
  for (int i = 0; i <= MAXL; i++) {
    int l = 0, r = min(n / max(i, 1), prv) + 1, m;
    while (l + 1 < r) {
      m = (l + r) / 2;
      if (get(m) <= i)
        r = m;
      else
        l = m;
    }
    for (int j = r; j < prv; j++) ans[j] = i;
    prv = r;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << "\n";
  return 0;
}
