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>
#pragma GCC optimize(2)
using namespace std;
int n, t, ver[100005 << 1], nex[100005 << 1], head[100005];
int tot, l, bound, ex[100005], dp[100005], copy_dp1, copy_dp[100005];
inline int read() {
  int s = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) s = (s << 1) + (s << 3) + ch - '0', ch = getchar();
  return s * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
void add_edge(int x, int y) {
  ver[++tot] = y;
  nex[tot] = head[x];
  head[x] = tot;
}
void dfs(int x, int f) {
  for (register int i = head[x]; i; i = nex[i]) {
    int y = ver[i];
    if (y == f) continue;
    dfs(y, x);
    dp[x] += dp[y];
  }
  int maxx = 0, secmaxx = 0;
  for (register int i = head[x]; i; i = nex[i]) {
    int y = ver[i];
    if (y == f) continue;
    if (ex[y] >= maxx) {
      secmaxx = maxx;
      maxx = ex[y];
    } else if (maxx > ex[y] && ex[y] >= secmaxx)
      secmaxx = ex[y];
  }
  if (maxx + secmaxx + 1 >= bound)
    dp[x]++;
  else
    ex[x] = maxx + 1;
}
int main() {
  n = read();
  t = (int)(log(n) / log(2)) + 1;
  for (register int i = 1; i < n; i++) {
    int x = read(), y = read();
    add_edge(x, y);
    add_edge(y, x);
  }
  memset(copy_dp, -1, sizeof copy_dp);
  for (l = 1; l <= n; l = bound + 1) {
    bound = l;
    if (copy_dp[bound] == -1) {
      memset(dp, 0, sizeof dp);
      memset(ex, 0, sizeof ex);
      dfs(1, 0);
      copy_dp[bound] = dp[1];
    }
    dp[1] = copy_dp[bound];
    copy_dp1 = dp[1];
    bound = 0;
    for (register int i = t; i >= 0; i--)
      if (bound + (1 << i) <= n) {
        bound += (1 << i);
        if (bound <= l) continue;
        if (copy_dp[bound] == -1) {
          memset(dp, 0, sizeof dp);
          memset(ex, 0, sizeof ex);
          dfs(1, 0);
          copy_dp[bound] = dp[1];
        }
        dp[1] = copy_dp[bound];
        if (dp[1] < copy_dp1) bound -= (1 << i);
      }
    for (register int i = l; i <= bound; i++) write(copy_dp1), putchar('\n');
  }
}
