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 = 100100;
const double eps = 1e-6;
vector<int> e[maxn], g[maxn];
int f[maxn], mx[maxn], ans[maxn];
int id[maxn], vid[maxn], fa[maxn], dfn, n;
void dfs(int u, int ff) {
  vid[u] = dfn, id[dfn--] = u, fa[u] = ff;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == ff) continue;
    dfs(v, u);
    g[vid[u]].push_back(vid[v]);
  }
}
void dp(int k) {
  for (int u = 1; u <= n; u++) {
    f[u] = 0;
    int mx1 = 0, mx2 = 0;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      f[u] += f[v];
      if (mx[v] >= mx1)
        mx2 = mx1, mx1 = mx[v];
      else if (mx[v] >= mx2)
        mx2 = mx[v];
    }
    if (mx1 + mx2 + 1 >= k)
      f[u]++, mx[u] = 0;
    else
      mx[u] = mx1 + 1;
  }
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfn = n;
  int rt = 1;
  dfs(rt, 0);
  int S = min(n, 800);
  for (int i = 1; i <= S; i++) {
    dp(i);
    ans[i] = f[n];
  }
  int ret, L, R, pre = S;
  for (int i = ans[S]; i >= 1; i--) {
    L = S, R = n, ret = 0;
    while (L <= R) {
      int mid = (L + R) / 2;
      dp(mid);
      if (f[n] >= i)
        ret = max(ret, mid), L = mid + 1;
      else
        R = mid - 1;
    }
    if (ret > pre) {
      for (int j = pre + 1; j <= ret; j++) ans[j] = i;
      pre = ret;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
