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 = 100010;
int n, s, tt, tp;
int head[N], to[N * 2], nxt[N * 2], f[N], id[N], v[N];
inline int gi() {
  int x = 0, o = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') o = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * o;
}
inline void dfs(int x, int fa) {
  v[++tp] = x, id[x] = tp;
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != fa) dfs(to[i], x);
}
inline int F(int k) {
  int ret = 0;
  for (int i = n; i; i--) {
    int x = v[i], A = 0, B = 0;
    for (int j = head[x]; j; j = nxt[j]) {
      int y = to[j];
      if (id[y] > id[x] && f[y] > B) {
        if (f[y] > A)
          B = A, A = f[y];
        else
          B = f[y];
      }
    }
    if (A + B + 1 >= k)
      ++ret, f[x] = 0;
    else
      f[x] = A + 1;
  }
  return ret;
}
int main() {
  cin >> n;
  s = sqrt(n) + 1e-9;
  for (int i = 1; i < n; i++) {
    int x = gi(), y = gi();
    to[++tt] = y, nxt[tt] = head[x], head[x] = tt;
    to[++tt] = x, nxt[tt] = head[y], head[y] = tt;
  }
  dfs(1, 0);
  for (int i = 1; i <= s; i++) printf("%d\n", F(i));
  for (int i = s + 1; i <= n;) {
    int k = F(i), l = i + 1, r = n;
    if (k) {
      r = n / k;
      while (l <= r) {
        int mid = (l + r) >> 1;
        if (F(mid) == k)
          l = mid + 1;
        else
          r = mid - 1;
      }
    }
    while (i <= r) printf("%d\n", k), ++i;
  }
  return 0;
}
