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 = 100005;
int n, ans[N], S, f[N], p[N], idx, dp[N];
vector<int> v[N];
void dfs(int pos, int fa) {
  f[pos] = fa;
  for (auto &i : v[pos])
    if (i != fa) dfs(i, pos);
  p[++idx] = pos;
}
int solve(int kk) {
  for (int i = 1; i <= n; i++) dp[i] = 1;
  int ret = 0;
  for (int i = 1; i <= n; i++) {
    int pos = p[i];
    if (f[pos] && dp[pos] != -1 && dp[f[pos]] != -1) {
      if (dp[pos] + dp[f[pos]] >= kk)
        ++ret, dp[f[pos]] = -1;
      else
        dp[f[pos]] = max(dp[f[pos]], dp[pos] + 1);
    }
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  S = sqrt(n * log2(n));
  int t1, t2;
  for (int i = 1; i < n; i++) {
    cin >> t1 >> t2;
    v[t1].push_back(t2);
    v[t2].push_back(t1);
  }
  dfs(1, 0);
  ans[1] = n;
  for (int i = 2; i <= S; i++) ans[i] = solve(i);
  int pos;
  for (int i = S + 1; i <= n; i = pos + 1) {
    int l = i, r = n, cur = solve(i);
    pos = -1;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (solve(mid) == cur)
        pos = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    for (int j = i; j <= pos; j++) ans[j] = cur;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << endl;
  return 0;
}
