Problem: Let G be a simple graph. Let W be a non-empty subset of vertices. Then W is almost-k-uniform if for each pair of distinct vertices u,v ∈ W the distance between u and v is either k or k+1.

You are given a tree on n vertices. For each i between 1 and n, find the maximum size of an almost-i-uniform set.

Input

The first line contains a single integer n (2 ≤ n ≤ 5 ⋅ 10^5) – the number of vertices of the tree.

Then n-1 lines follows, the i-th of which consisting of two space separated integers u_i, v_i (1 ≤ u_i, v_i ≤ n) meaning that there is an edge between vertices u_i and v_i. 

It is guaranteed that the given graph is tree. 

Output

Output a single line containing n space separated integers a_i, where a_i is the maximum size of an almost-i-uniform set.

Examples

Input


5
1 2
1 3
1 4
4 5


Output


4 3 2 1 1


Input


6
1 2
1 3
1 4
4 5
4 6


Output


4 4 2 1 1 1

Note

Consider the first example. 

  * The only maximum almost-1-uniform set is \{1, 2, 3, 4\}. 
  * One of the maximum almost-2-uniform sets is or \{2, 3, 5\}, another one is \{2, 3, 4\}. 
  * A maximum almost-3-uniform set is any pair of vertices on distance 3. 
  * Any single vertex is an almost-k-uniform set for k ≥ 1. 



In the second sample there is an almost-2-uniform set of size 4, and that is \{2, 3, 5, 6\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MX = 5e5 + 1;
int cnt[MX];
int oth[MX];
int best[MX];
int ans[MX * 2];
vector<int> adj[MX];
vector<int> deps[MX];
void dfs_down(int u, int p) {
  for (auto c : adj[u]) {
    if (c != p) {
      dfs_down(c, u);
      if (best[c] + 1 > best[u]) {
        oth[u] = best[u];
        best[u] = best[c] + 1;
      } else if (best[c] + 1 > oth[u]) {
        oth[u] = best[c] + 1;
      }
    }
  }
  deps[p].push_back(best[u] + 1);
}
void dfs_up(int u, int p) {
  if (p) {
    deps[u].push_back(best[p] + 1);
    if (best[p] + 1 > best[u]) {
      oth[u] = best[u];
      best[u] = best[p] + 1;
    } else if (best[p] + 1 > oth[u]) {
      oth[u] = best[p] + 1;
    }
  }
  for (auto c : adj[u]) {
    if (c != p) {
      if (best[u] == best[c] + 1) {
        swap(best[u], oth[u]);
      }
      dfs_up(c, u);
      if (oth[u] > best[u]) {
        swap(oth[u], best[u]);
      }
    }
  }
}
void dfs(int u, int p) {
  vector<pair<int, int>> chd;
  sort(deps[u].rbegin(), deps[u].rend());
  for (int i = 0; i < int(deps[u].size()); ++i) {
    int d = deps[u][i];
    chd.emplace_back(d, u);
    if (i && d + 1 <= deps[u][i - 1]) {
      ans[d * 2 + 1] = max(ans[d * 2 + 1], i + 1);
    }
    ans[d * 2 - 1] = max(ans[d * 2 - 1], i + 1);
    ans[d * 2] = max(ans[d * 2], i + 1);
  }
  for (auto c : adj[u]) {
    if (c != p) {
      dfs(c, u);
      cnt[c] = 0;
      for (auto i : deps[c]) {
        chd.emplace_back(i, c);
      }
    }
  }
  int mx = 0;
  sort(chd.rbegin(), chd.rend());
  for (auto [d, i] : chd) {
    ++cnt[i];
    if (i != u) {
      mx = max(mx, cnt[i]);
    }
    ans[2 * d] = max(ans[2 * d], cnt[u] + mx - 2);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  fill(ans, ans + n + 1, 1);
  for (int i = 1; i <= n; ++i) {
    ans[1] = max(ans[1], int(adj[i].size()) + 1);
  }
  dfs_down(1, 0);
  dfs_up(1, 0);
  dfs(1, 0);
  for (int i = n - 2; i; --i) {
    ans[i] = max(ans[i], ans[i + 2]);
  }
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}