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<int> adj[300005];
int par[300005], lvl[300005];
void dfs(int n, int pr) {
  par[n] = pr;
  lvl[n] = lvl[pr] + 1;
  for (auto &cv : adj[n]) {
    if (cv == pr) continue;
    dfs(cv, n);
  }
}
vector<int> v;
pair<int, int> pr[300005];
int sh(int k) {
  for (auto &n : v) {
    pr[n].first = 0;
    int mx1, mx2;
    mx1 = mx2 = 0;
    for (auto &cv : adj[n]) {
      if (cv == par[n]) continue;
      pr[n].first += pr[cv].first;
      if (pr[cv].second >= mx1) {
        mx2 = mx1;
        mx1 = pr[cv].second;
      } else if (pr[cv].second > mx2) {
        mx2 = pr[cv].second;
      }
    }
    if (mx2 + mx1 + 1 >= k) {
      pr[n].first++;
      pr[n].second = 0;
    } else {
      pr[n].second = mx1 + 1;
    }
  }
  return pr[1].first;
}
bool cmp(int x, int y) {
  if (lvl[x] != lvl[y])
    return lvl[x] > lvl[y];
  else
    return x < y;
}
int ans[300005], lg[300005];
int32_t main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) v.push_back(i);
  for (int i = 0; i <= 17; i++) lg[1 << i] = i;
  for (int i = 200000; i >= 0; i--)
    if (lg[i])
      continue;
    else
      lg[i] = lg[i + 1];
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  sort(v.begin(), v.end(), cmp);
  int l = n + 1;
  int pre = 0;
  int x = sqrt(n);
  for (int i = n; i >= 1; i--) {
    if (l <= i) {
      ans[i] = pre;
      continue;
    }
    if (i < x) {
      ans[i] = sh(i);
      continue;
    }
    pre = sh(i);
    l = i;
    int flag = 0;
    for (int j = lg[i]; j >= 0; j--) {
      if (l - (1 << j) >= 1 && sh(l - (1 << j)) == pre) l -= 1 << j;
    }
    ans[i] = pre;
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
