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 N = 1e5 + 5;
struct node {
  int to, nxt;
} edge[N << 1];
int n, num, head[N], ans[N], res, f[N];
inline int read() {
  int x = 0, w = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') c = getchar(), w = -1;
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * w;
}
inline void add_edge(int from, int to) {
  edge[++num] = (node){to, head[from]};
  head[from] = num;
}
void dfs(int k, int fa, int x) {
  int fir = 0, sec = 0;
  for (register int i = head[k]; i; i = edge[i].nxt) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, k, x);
    if (f[v] > fir)
      sec = fir, fir = f[v];
    else
      sec = max(sec, f[v]);
  }
  if (fir + sec + 1 >= x)
    f[k] = 0, ++res;
  else
    f[k] = fir + 1;
}
void solve(int s, int t, int l, int r) {
  if (s > t || l > r) return;
  if (l == r) {
    for (register int i = s; i <= t; i++) ans[i] = l;
    return;
  }
  int mid = (s + t) >> 1;
  res = 0;
  dfs(1, 0, mid);
  ans[mid] = res;
  solve(s, mid - 1, res, r);
  solve(mid + 1, t, l, res);
}
int main() {
  n = read();
  for (register int i = 1; i < n; i++) {
    int x = read(), y = read();
    add_edge(x, y);
    add_edge(y, x);
  }
  solve(1, n, 0, n);
  for (register int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
