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 = 1e5 + 5, M = N << 1;
int n, m, tot, tim, cnt, head[N], nex[M], to[M], fa[N], dfn[N], ans[N], f[N];
void add(int x, int y) { to[++tot] = y, nex[tot] = head[x], head[x] = tot; }
void dfs(int x, int ff) {
  fa[x] = ff;
  int ver, i;
  for (i = head[x]; i; i = nex[i]) {
    ver = to[i];
    if (ver == ff) continue;
    dfs(ver, x);
  }
  dfn[++tim] = x;
}
int solve(int x) {
  int res = 0, i, ver, ff;
  for (i = 1; i <= n; i++) f[i] = 1;
  for (i = 1; i <= n; i++) {
    ver = dfn[i], ff = fa[ver];
    if (ff && f[ff] != -1 && f[ver] != -1) {
      if (f[ver] + f[ff] >= x)
        res++, f[ff] = -1;
      else
        f[ff] = max(f[ff], f[ver] + 1);
    }
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int i, u, v, l, r, num, tmp, mid;
  cnt = sqrt(n * log2(n));
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    add(u, v), add(v, u);
  }
  ans[1] = n, dfs(1, 0);
  for (i = 2; i <= cnt; i++) ans[i] = solve(i);
  for (i = cnt + 1; i <= n; i++) {
    num = solve(i), l = i, r = n, tmp = i;
    while (l + 1 < r) {
      mid = l + r >> 1;
      if (solve(mid) == num)
        tmp = max(tmp, mid), l = mid;
      else
        r = mid;
    }
    for (; i <= tmp; i++) ans[i] = num;
    i--;
  }
  for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
