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>
inline int gi() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int fir[100010], dis[200010], nxt[200010], id;
inline void link(int a, int b) { nxt[++id] = fir[a], fir[a] = id, dis[id] = b; }
std::vector<int> G[100010];
int tot, mid, d;
inline int dfs(int x) {
  int mx = 0, cmx = 0;
  for (int i = 0; i < G[x].size(); ++i) {
    d = dfs(G[x][i]);
    if (d > mx)
      cmx = mx, mx = d;
    else if (d > cmx)
      cmx = d;
  }
  if (mx + cmx + 1 >= mid) ++tot, mx = -1;
  return mx + 1;
}
inline void DFS(int x, int fa = -1) {
  for (int i = fir[x]; i; i = nxt[i]) {
    if (fa == dis[i]) continue;
    G[x].push_back(dis[i]);
    DFS(dis[i], x);
  }
}
int R[100010];
int main() {
  int n = gi(), a, b, blk = sqrt(n * log2(n)), lim = n / blk, l, r;
  for (int i = 1; i < n; ++i) a = gi(), b = gi(), link(a, b), link(b, a);
  DFS(1);
  for (int i = 1; i <= blk; ++i) mid = i, tot = 0, dfs(1), printf("%d\n", tot);
  for (int i = 0; i <= lim; ++i) {
    l = 0, r = n;
    while (l < r) {
      mid = ((l + r) >> 1) + 1;
      tot = 0;
      dfs(1);
      if (tot >= i)
        l = mid;
      else
        r = mid - 1;
    }
    R[i] = l;
  }
  int p = lim;
  for (int i = blk + 1; i <= n; ++i) {
    while (R[p] < i) --p;
    printf("%d\n", p);
  }
  return 0;
}
