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;
vector<int> g[100500];
int p[100500], f[100500], tn;
void pfs(int u, int fa) {
  f[u] = fa;
  for (int i = 0; i < g[u].size(); i++)
    if (g[u][i] != fa) pfs(g[u][i], u);
  p[++tn] = u;
}
int n, mx[100500];
int calc(int lim) {
  int cnt = 0;
  for (int t = 1; t <= n; t++) {
    int u = p[t], l0 = 0, l1 = 0;
    for (int i = 0; i < g[u].size(); i++)
      if (f[u] != g[u][i]) {
        int sav = mx[g[u][i]];
        if (sav > l0) {
          l1 = l0;
          l0 = sav;
        } else
          l1 = max(l1, sav);
      }
    if (l0 + l1 + 1 >= lim) {
      cnt++;
      mx[u] = 0;
    } else
      mx[u] = l0 + 1;
  }
  return cnt;
}
int nxt(int l, int c) {
  int r = n, mid;
  while (l < r) {
    mid = (l + r + 1) >> 1;
    if (calc(mid) == c)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int BS;
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  pfs(1, 0);
  BS = min((int)sqrt(n * log(n)), n);
  for (int i = 1; i <= BS; i++) printf("%d ", calc(i));
  for (int i = BS + 1; i <= n;) {
    int sav = calc(i), p = nxt(i, sav);
    for (int j = i; j <= p; j++) printf("%d ", sav);
    i = p + 1;
  }
  return 0;
}
