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;
int n, q, x, y, head[200002], Fa[100001], dfsx[100001], f[100001],
    cnt = 0, tot = 0, t, l, r, mid;
struct node {
  int nxt, to;
} e[200002];
void add(int x, int y) {
  e[++cnt].nxt = head[x];
  e[cnt].to = y;
  head[x] = cnt;
}
void dfs(int x, int fa) {
  Fa[x] = fa;
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v != fa) dfs(v, x);
  }
  dfsx[++tot] = x;
  return;
}
int solve(int x) {
  int k, ans = 0;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) {
    int k = dfsx[i];
    if (Fa[k] && ~f[Fa[k]] && ~f[k]) {
      if (f[k] + f[Fa[k]] >= x) {
        f[Fa[k]] = -1;
        ans++;
      } else
        f[Fa[k]] = max(f[Fa[k]], f[k] + 1);
    }
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  q = sqrt(n * log(n) / log(2));
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  printf("%d\n", n);
  for (int i = 2; i <= q; i++) printf("%d\n", solve(i));
  for (int i = q + 1; i <= n; i = l + 1) {
    l = i;
    r = n;
    t = solve(i);
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (solve(mid) == t)
        l = mid;
      else
        r = mid - 1;
    }
    for (int j = i; j <= l; j++) printf("%d\n", t);
  }
  return 0;
}
