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>
const int N = 100001;
int n;
int answer[N];
int d[N];
int calc[N];
int fa[N];
int id[N];
int fafafa;
struct edge {
  int to;
  int nxt;
} e[N << 1];
int head[N], tot;
void add(int x, int y) {
  e[++tot].to = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void DFS(int x, int f) {
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == f) continue;
    fa[y] = x;
    DFS(y, x);
  }
  id[++fafafa] = x;
}
int dfs(int L) {
  int ans = 0;
  for (int i = 1; i <= n; ++i) d[i] = 1;
  for (int i = 1; i <= n; ++i) {
    int x = id[i];
    if (fa[x] && (d[x]) && ~(d[fa[x]])) {
      if (d[x] + d[fa[x]] >= L) {
        ++ans;
        d[fa[x]] = -1;
      } else {
        d[fa[x]] = std::max(d[fa[x]], d[x] + 1);
      }
    }
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  DFS(1, 0);
  int size = sqrt(n * (log(n) / log(2)));
  answer[1] = n;
  for (int i = 2; i <= size; ++i) answer[i] = dfs(i);
  for (int i = size + 1, l, r; i <= n; i = l + 1) {
    l = i, r = n;
    int now = calc[i] ? calc[i] : calc[i] = dfs(i);
    while (l + 1 < r) {
      int mid = l + r >> 1;
      if ((calc[mid] ? calc[mid] : calc[mid] = dfs(mid)) == now)
        l = mid;
      else
        r = mid;
    }
    for (int j = i; j <= l; ++j) answer[j] = now;
  }
  for (int i = 1; i <= n; ++i) printf("%d ", answer[i]);
  return 0;
}
