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>
struct edge {
  int to, nxt;
} e[100005 << 1];
int nowk, num[100005], len[100005], head[100005], tot, ans[100005], n,
    f[100005], dep[100005], rk[100005], max[100005], smax[100005];
inline void addedge(int x, int y) {
  e[++tot] = (edge){y, head[x]};
  head[x] = tot;
  e[++tot] = (edge){x, head[y]};
  head[y] = tot;
}
void dfs(int x) {
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != f[x])
      dep[e[i].to] = dep[x] + 1, f[e[i].to] = x, dfs(e[i].to);
}
inline bool cmp(register int i, register int j) { return dep[i] < dep[j]; }
inline void init() {
  for (register int i = 1; i <= n; i++) rk[i] = i;
  std::sort(rk + 1, rk + n + 1, cmp);
}
inline int solve() {
  for (register int i = 1; i <= n; i++) num[i] = max[i] = smax[i] = 0;
  for (register int i = n; i; i--) {
    register int now = rk[i], f = ::f[now];
    if (max[now] + smax[now] + 1 >= nowk)
      len[now] = 0, num[now]++;
    else
      len[now] = max[now] + 1;
    if (!f) break;
    num[f] += num[now];
    if (len[now] >= max[f])
      smax[f] = max[f], max[f] = len[now];
    else if (len[now] > smax[f])
      smax[f] = len[now];
  }
  return num[1];
}
inline int solve(int k) {
  if (~ans[k]) return ans[k];
  nowk = k;
  return ans[k] = solve();
}
int main() {
  std::scanf("%d", &n);
  for (int i = 1; i <= n; i++) ans[i] = -1;
  for (int i = 1, x, y; i < n; i++) std::scanf("%d%d", &x, &y), addedge(x, y);
  dfs(1);
  init();
  int blo = sqrt(n);
  register int now = 1, nowans = solve(1), end = solve(n);
  while (now <= n) {
    register int l = now, r = std::min(n, l + blo), ans;
    if (nowans == end) {
      for (register int i = now; i <= n; ++i) printf("%d\n", nowans);
      return 0;
    }
    while (l <= r) {
      register int mid = l + r >> 1;
      if (solve(mid) == nowans)
        l = mid + 1, ans = mid;
      else
        r = mid - 1;
    }
    for (register int i = now; i <= ans; i++) std::printf("%d\n", nowans);
    now = ans + 1;
    if (now <= n) nowans = solve(now);
  }
  return 0;
}
