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 = 620;
const long long LINF = 1e18;
const int LOG = 31;
const int INF = 1e9;
const int P = 31;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int SQ = 350;
const int MOD = 1e9 + 7;
vector<int> ed[N], adj[N];
int n, dp1[N], dp2[N], ans[N], vis[N], t[N], tim;
void dfs(int x) {
  if (vis[x]) return;
  vis[x] = 1;
  for (int i : ed[x])
    if (vis[i] == 0) dfs(i);
  t[x] = ++tim;
}
int calc(int k) {
  for (int cur = 1; cur <= n; cur++) {
    dp1[cur] = dp2[cur] = 0;
    int mx2 = 0, mx1 = 0;
    for (auto i : adj[cur]) {
      dp1[cur] += dp1[i];
      mx2 = max(mx2, dp2[i]);
      if (mx1 < mx2) swap(mx1, mx2);
    }
    if (mx1 + mx2 + 1 >= k)
      dp1[cur]++;
    else
      dp2[cur] = mx1 + 1;
  }
  return dp1[n];
}
void solve(int l, int r, int ansl, int ansr) {
  if (l > r || ansl > ansr) return;
  if (ansl == ansr)
    ans[(l + r) / 2] = ansl;
  else
    ans[(l + r) / 2] = calc((l + r) / 2);
  solve(l, (l + r) / 2 - 1, ans[(l + r) / 2], ansr);
  solve((l + r) / 2 + 1, r, ansl, ans[(l + r) / 2]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    for (int j : ed[i]) {
      if (t[i] < t[j]) continue;
      adj[t[i]].push_back(t[j]);
    }
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
}
