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;
vector<int> adj[100010];
pair<int, int> dp[100010];
int ans[100010], val[100010];
int dep[100010], fa[100010];
int id[100010], n;
void pre_dfs(int u, int ma = 0) {
  fa[u] = ma;
  dep[u] = dep[ma] + 1;
  for (int v : adj[u]) {
    if (v != ma) pre_dfs(v, u);
  }
}
void solve(int first) {
  for (int i = 1; i <= n; i++) {
    int u = id[i];
    if (first == 1)
      dp[u] = {1, 0};
    else
      dp[u] = {0, 1};
    for (int v : adj[u]) {
      if (v == fa[u]) continue;
      if (dp[u].second + dp[v].second >= first) {
        dp[u] = {dp[u].first + dp[v].first + 1, 0};
      } else {
        int d = 0;
        if (dp[u].second) d = max(1 + dp[v].second, dp[u].second);
        dp[u] = {dp[u].first + dp[v].first, d};
      }
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(val, -1, sizeof val);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  pre_dfs(1);
  for (int i = 1; i <= n; i++) id[i] = i;
  sort(id + 1, id + n + 1, [&](int i, int j) { return dep[i] > dep[j]; });
  int N = min(1000, n);
  for (int i = 1; i <= N; i++) {
    if (val[i] == -1) {
      solve(i);
      val[i] = dp[1].first;
    }
    ans[i] = val[i];
  }
  for (int i = 1; i <= n / (N + 1); i++) {
    int st = 0, ed = n / i + 1;
    while (st + 1 < ed) {
      int mid = (st + ed) >> 1;
      if (val[mid] == -1) {
        solve(mid);
        val[mid] = dp[1].first;
      }
      if (val[mid] >= i)
        st = mid;
      else
        ed = mid;
    }
    if (st > N) ans[st] = i;
  }
  for (int i = n; i > N; i--) ans[i] = max(ans[i], ans[i + 1]);
  for (int i = 1; i <= n; i++) cout << ans[i] << endl;
}
