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;
template <typename T>
inline void read(T& t) {
  T f = 1;
  char ch = getchar();
  t = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') t = t * 10 + ch - '0', ch = getchar();
  t *= f;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
int n;
int head[100005], ne[200005], to[200005], total = 1;
inline void add(int a, int b) {
  total++;
  to[total] = b;
  ne[total] = head[a];
  head[a] = total;
}
int id[200005], cnt, fa[200005];
void dfs(int now, int fas) {
  for (int i = head[now]; i; i = ne[i])
    if (to[i] != fas) dfs(to[i], now);
  id[++cnt] = now;
  fa[now] = fas;
}
int f[200005];
long long solve(int x) {
  int now, re = 0;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) {
    now = id[i];
    if (fa[now] && f[fa[now]] != -1 && f[now] != -1)
      if (f[fa[now]] + f[now] >= x)
        re++, f[fa[now]] = f[now] = -1;
      else
        f[fa[now]] = max(f[fa[now]], f[now] + 1);
  }
  return re;
}
int main() {
  read(n);
  for (int i = 1, x, y; i < n; i++) read(x, y), add(x, y), add(y, x);
  dfs(1, 0);
  int q = sqrt(n * (log(n) / log(2)));
  printf("%d\n", n);
  for (int i = 2; i <= q; i++) printf("%lld\n", solve(i));
  for (int i = q + 1, l, r, mid; i <= n; i = r + 1) {
    l = i;
    r = n;
    int tp = solve(i);
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (solve(mid) != tp)
        r = mid - 1;
      else
        l = mid;
    }
    for (int j = i; j <= l; j++) printf("%d\n", tp);
  }
  return 0;
}
