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;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int head[202002], to[202002], nxt[202022];
int tot;
void add(int x, int y) {
  tot++;
  to[tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
int f[202200], dis[220222];
int cnt, d;
vector<int> G[101001];
int dfs(int x, int mid) {
  int mx = 0, cmx = 0;
  for (int i = 0; i < G[x].size(); ++i) {
    d = dfs(G[x][i], mid);
    if (d > mx)
      cmx = mx, mx = d;
    else if (d > cmx)
      cmx = d;
  }
  if (mx + cmx + 1 >= mid) ++cnt, mx = -1;
  return mx + 1;
}
void DFS(int x, int fa = -1) {
  for (int i = head[x]; i; i = nxt[i]) {
    if (fa == to[i]) continue;
    G[x].push_back(to[i]);
    DFS(to[i], x);
  }
}
int a[2020202];
int main() {
  int n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  DFS(1);
  int blk = sqrt(n * log2(n));
  for (int i = 1; i <= blk; i++) {
    cnt = 0;
    dfs(1, i);
    printf("%d\n", cnt);
  }
  for (int i = blk + 1; i <= n;) {
    cnt = 0;
    dfs(1, i);
    int l = i, r = n, tmp = cnt, ans, mid;
    while (l <= r) {
      mid = (l + r) / 2;
      cnt = 0;
      dfs(1, mid);
      if (cnt == tmp)
        ans = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    int j = ans;
    for (int k = i; k <= j; k++) {
      a[k] = tmp;
    }
    i = j + 1;
    j = i;
  }
  for (int i = blk + 1; i <= n; i++) {
    printf("%d\n", a[i]);
  }
  return 0;
}
