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 N = 100005;
int n, cc, tot, ans[N], f[N], hd[N], nxt[N << 1], to[N << 1];
void add(int u, int v) {
  nxt[++cc] = hd[u];
  hd[u] = cc;
  to[cc] = v;
}
void dfs(int x, int p, int k) {
  int mx = 0, mx2 = 0;
  for (int i = hd[x]; i; i = nxt[i])
    if (to[i] != p) {
      int v = to[i];
      dfs(v, x, k);
      if (f[v] > mx)
        mx2 = mx, mx = f[v];
      else if (f[v] > mx2)
        mx2 = f[v];
    }
  if (mx + mx2 + 1 >= k)
    f[x] = 0, tot++;
  else
    f[x] = mx + 1;
}
void sol(int l, int r, int tl, int tr) {
  if (l > r || tl > tr) return;
  if (tl == tr) {
    for (int i = l; i <= r; i++) ans[i] = tl;
    return;
  }
  int mid = l + r >> 1;
  tot = 0;
  dfs(1, 0, mid);
  ans[mid] = tot;
  sol(l, mid - 1, tot, tr);
  sol(mid + 1, r, tl, tot);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) scanf("%d%d", &u, &v), add(u, v), add(v, u);
  sol(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
