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 dp[111111];
int xx[111111];
vector<int> e[111111];
int nowk;
void dfs(int now, int p) {
  int maxnum[2] = {0, 0}, cnt = 0;
  for (int k : e[now])
    if (k != p) {
      dfs(k, now), cnt += dp[k];
      int x = xx[k];
      for (int i = 0; i < 2; i++)
        if (maxnum[i] < x) swap(x, maxnum[i]);
    }
  if (maxnum[0] + maxnum[1] + 1 >= nowk) cnt++, maxnum[0] = -1;
  dp[now] = cnt;
  xx[now] = maxnum[0] + 1;
}
int res[111111];
void solve(int l, int r, int ql, int qr) {
  if (l > r) return;
  if (ql == qr) {
    for (int i = l; i < r + 1; i++) res[i] = ql;
    return;
  }
  int mid = (l + r) / 2;
  nowk = mid;
  dfs(1, 1);
  res[mid] = dp[1];
  solve(l, mid - 1, dp[1], qr), solve(mid + 1, r, ql, dp[1]);
}
int main() {
  int n, a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    e[a].push_back(b), e[b].push_back(a);
  }
  solve(1, n, 0, n);
  for (int i = 1; i < n + 1; i++) printf("%d\n", res[i]);
}
