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;
int n, hed[100010], cnt, ans[100010], dp[100010];
struct node {
  int to, next;
} a[200010];
inline int read() {
  int x = 0;
  char s = getchar();
  while (s > '9' || s < '0') s = getchar();
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x;
}
inline void add(int from, int to) {
  a[++cnt].to = to;
  a[cnt].next = hed[from];
  hed[from] = cnt;
}
inline void dfs(int x, int f, int s) {
  dp[x] = 0;
  int zui = 0, ci = 0;
  for (int i = hed[x]; i; i = a[i].next) {
    int y = a[i].to;
    if (y == f) continue;
    dfs(y, x, s);
    if (dp[y] > zui) {
      ci = zui;
      zui = dp[y];
    } else if (dp[y] > ci)
      ci = dp[y];
  }
  if (zui + ci + 1 >= s)
    ++cnt;
  else
    dp[x] = zui + 1;
}
inline void solve(int l, int r, int le, int ri) {
  if (l > r) return;
  if (le > ri) return;
  if (le == ri) {
    for (int i = l; i <= r; ++i) ans[i] = le;
    return;
  }
  int mid = (l + r) >> 1;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  solve(l, mid - 1, cnt, ri);
  solve(mid + 1, r, le, cnt);
}
int main() {
  n = read();
  for (int i = 1; i <= n - 1; ++i) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
