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 INF = 1e9 + 7, MAXN = 1e5 + 10, MAXM = 2e5 + 10;
int head[MAXN], sz, to[MAXM], nxt[MAXM];
inline void add(int x, int y) {
  nxt[++sz] = head[x], head[x] = sz, to[sz] = y;
  nxt[++sz] = head[y], head[y] = sz, to[sz] = x;
}
int N, maxa;
int fa[MAXN], id[MAXN], tc;
int ans[MAXN], f[MAXN];
void dfs(int x, int y) {
  fa[x] = y;
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != y) dfs(to[i], x);
  id[++tc] = x;
}
inline int solve(int k) {
  for (int i = 1; i <= N; i++) f[i] = 1;
  int ret = 0;
  for (int i = 1; i <= N; i++) {
    int x = id[i];
    if (fa[x] && f[fa[x]] != -1 && f[x] != -1) {
      if (f[x] + f[fa[x]] >= k) {
        ret++;
        f[fa[x]] = -1;
      } else
        f[fa[x]] = max(f[fa[x]], f[x] + 1);
    }
  }
  return ret;
}
int main() {
  scanf("%d", &N);
  maxa = sqrt(N * log(N) / log(2));
  for (int i = 1; i < N; i++) {
    int ii, jj;
    scanf("%d%d", &ii, &jj);
    add(ii, jj);
  }
  ans[1] = N;
  dfs(1, 0);
  for (int i = 2; i <= maxa; i++) ans[i] = solve(i);
  for (int i = maxa + 1; i <= N;) {
    int l = i, r = N, check = solve(i);
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (solve(mid) != check)
        r = mid - 1;
      else
        l = mid;
    }
    for (int j = i; j <= l; j++) ans[j] = check;
    i = l + 1;
  }
  for (int i = 1; i <= N; i++) printf("%d\n", ans[i]);
  return 0;
}
