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;
vector<vector<int> > g, g1;
vector<int> d, h;
vector<int> ord;
void dfs0(int v, int pr) {
  for (int nv : g[v])
    if (nv != pr) {
      dfs0(nv, v);
      g1[v].push_back(nv);
    }
  ord.push_back(v);
}
void dfs(int v, int k) {
  d[v] = 0;
  h[v] = 1;
  pair<int, int> p(0, 0);
  for (int nv : g1[v]) {
    d[v] += d[nv];
    if (p.second <= h[nv]) {
      p.first = p.second;
      p.second = h[nv];
    } else
      p.first = max(p.first, h[nv]);
  }
  if (p.first + p.second + 1 >= k) {
    h[v] = 0;
    d[v]++;
  } else {
    h[v] = 1 + p.second;
  }
}
map<int, int> saved;
int solve(int k) {
  if (saved.count(k)) return saved[k];
  d.resize(g.size());
  h.resize(g.size());
  for (int v : ord) dfs(v, k);
  saved[k] = d[0];
  return d[0];
}
void run(int l, int r, vector<int>& res) {
  if (r - l < 2) return;
  if (res[l] == res[r]) {
    for (int i = l + 1; i < r; ++i) res[i] = res[l];
    return;
  }
  int m = (l + r) / 2;
  res[m] = solve(m);
  run(l, m, res);
  run(m, r, res);
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  int n = 100000;
  cin >> n;
  g.resize(n);
  g1.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a = i + 1, b = i + 2;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs0(0, -1);
  vector<int> res(n + 2, -1);
  res[1] = n;
  res[n + 1] = 0;
  run(1, n + 1, res);
  for (int i = 1; i <= n; ++i) cout << res[i] << '\n';
  return 0;
}
