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 MAXN = 2e5 + 21;
int n, x, y, res;
vector<int> g[MAXN];
int f[MAXN], ans[MAXN];
void dfs(int u, int k, int fa) {
  int mx = 0, mx2 = 0;
  for (auto v : g[u]) {
    if (v == fa) continue;
    dfs(v, k, u);
    if (f[v] > mx) {
      mx2 = mx;
      mx = f[v];
    } else
      mx2 = max(mx2, f[v]);
  }
  if (mx + mx2 + 1 >= k) {
    res++;
    f[u] = 0;
  } else
    f[u] = mx + 1;
}
void solve(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = (l + r) / 2;
  res = 0;
  dfs(1, mid, 0);
  ans[mid] = res;
  solve(l, mid - 1, res, R);
  solve(mid + 1, r, L, res);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) cout << ans[i] << '\n';
  return 0;
}
