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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops,no-stack-protector,fast-math")
using namespace std;
const int MAXN = 500000;
vector<int> oldadj[MAXN];
vector<int> adj[MAXN];
int k;
int tin[MAXN];
int curtm;
void dfs1(int u, int p) {
  tin[u] = curtm++;
  for (int v : oldadj[u]) {
    if (v != p) {
      dfs1(v, u);
    }
  }
}
pair<int, int> dfs(int u, int p) {
  int pref = 0;
  int min_smallest_out = 1000000000;
  int premin_smallest_out = 1000000000;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) {
      continue;
    }
    pair<int, int> res = dfs(v, u);
    int lrg = res.first;
    int out = res.second;
    pref += lrg;
    if (out > 0) {
      if (out <= min_smallest_out) {
        premin_smallest_out = min_smallest_out;
        min_smallest_out = out;
      } else if (out <= premin_smallest_out) {
        premin_smallest_out = out;
      }
    }
  }
  int sml;
  if (premin_smallest_out < 1000000000 &&
      min_smallest_out - 1 <= k - premin_smallest_out) {
    sml = 0;
  } else {
    sml = min(k, min_smallest_out) - 1;
  }
  return make_pair(pref + (sml == 0), sml);
}
int go(int i) {
  k = i;
  return dfs(0, -1).first;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    oldadj[u].push_back(v);
    oldadj[v].push_back(u);
  }
  dfs1(0, -1);
  for (int u = 0; u < n; u++) {
    for (int v : oldadj[u]) {
      adj[tin[u]].push_back(tin[v]);
    }
  }
  vector<int> ans(n + 1);
  vector<pair<int, int> > bounds;
  for (int h = 100; h >= 0; h--) {
    int l = 0;
    int r = n + 1;
    while (r - l > 1) {
      int m = (l + r) / 2;
      if (go(m) > h) {
        l = m;
      } else {
        r = m;
      }
    }
    bounds.emplace_back(r, h);
  }
  for (int i = 0; i < bounds.size() - 1; i++) {
    for (int j = bounds[i].first; j < bounds[i + 1].first; j++) {
      ans[j] = bounds[i].second;
    }
  }
  for (int i = 1; i <= n; i++) {
    ans[i] = go(i);
    if (ans[i] < 100) {
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", ans[i]);
  }
  printf("\n");
  return 0;
}
