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 std::cerr;
using std::cin;
using std::max;
using std::min;
int n, a, b, ecnt, last[100005], tot, f[100005], ans[100005];
struct road {
  int to, nex;
} e[100005 << 1];
void adde(int u, int v) { e[++ecnt] = (road){v, last[u]}, last[u] = ecnt; }
void dfs(int x, int fath, int v) {
  int mx1 = 0, mx2 = 0;
  for (int k = last[x]; k; k = e[k].nex)
    if (e[k].to ^ fath) {
      dfs(e[k].to, x, v);
      if (f[e[k].to] > mx1)
        mx2 = mx1, mx1 = f[e[k].to];
      else if (f[e[k].to] > mx2)
        mx2 = f[e[k].to];
    }
  if (mx1 + mx2 + 1 >= v)
    tot++, f[x] = 0;
  else
    f[x] = mx1 + 1;
}
void Solve(int l, int r, int ql, int qr) {
  if (ql > qr) return;
  if (l == r) {
    for (int i = ql; i <= qr; i++) ans[i] = l;
    return;
  }
  int mid = ql + qr >> 1;
  tot = 0, dfs(1, 0, mid), ans[mid] = tot;
  Solve(l, tot, mid + 1, qr), Solve(tot, r, ql, mid - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d", &a, &b), adde(a, b), adde(b, a);
  Solve(0, n, 1, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
