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;
const int block = 1000;
const int N = 2e5;
int used[N], lnk[N], hg[N], k;
vector<int> ed[N];
vector<int> vr[N];
int solve(int v, int p, int b) {
  int mx1 = 0, mx2 = 0;
  used[v] = 0;
  int ans = 0;
  for (auto to : ed[v]) {
    if (to == p) continue;
    ans += solve(to, v, b);
    if (used[to]) continue;
    if (hg[to] > mx1) {
      mx2 = mx1;
      mx1 = hg[to];
    } else
      mx2 = max(mx2, hg[to]);
  }
  if (mx1 + mx2 + 1 >= b) used[v] = 1, ans++;
  hg[v] = mx1 + 1;
  return ans;
}
void dfs(int v, int p) {
  k++;
  lnk[v] = k;
  for (auto to : vr[v]) {
    if (to == p) continue;
    dfs(to, v);
    ed[lnk[v]].push_back(lnk[to]);
    ed[lnk[to]].push_back(lnk[v]);
  }
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(0);
  cin >> n;
  n++;
  n--;
  n--;
  n++;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    vr[u].push_back(v);
    vr[v].push_back(u);
  }
  k = 0;
  dfs(1, 0);
  for (int i = 1; i <= min(block, n); i++) {
    cout << solve(1, 0, i) << "\n";
  }
  int i = block + 1;
  while (i <= n) {
    int l = i;
    int r = n;
    int ans = solve(1, 0, l);
    while (r - l > 0) {
      int mid = (r + l) / 2 + 1;
      if (solve(1, 0, mid) == ans)
        l = mid;
      else
        r = mid - 1;
    }
    for (; i <= r; i++) cout << ans << "\n";
  }
  return 0;
}
