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 mod1 = 998244353, mod2 = 1e9 + 7;
const int MAX_N = 100005, MAX_M = 200010;
vector<int> grp[MAX_N];
pair<int, int> dp[MAX_N];
int par[MAX_N], dp_ord[MAX_N], ans[MAX_N], max1[MAX_N], max2[MAX_N], n, c;
void dfs(int v, int p) {
  par[v] = p;
  for (auto u : grp[v]) {
    if (u != p) {
      dfs(u, v);
    }
  }
  dp_ord[c++] = v;
}
int calc(int k) {
  for (int i = 0; i < n; i++)
    dp[i].first = dp[i].second = max1[i] = max2[i] = 0;
  for (int i = 0; i < n; i++) {
    int t = dp_ord[i];
    if (max1[t] + max2[t] + 1 >= k) {
      dp[t].first = max1[t] + max2[t] + 1;
    } else {
      dp[t].first = max1[t] + 1;
    }
    if (dp[t].first >= k) {
      dp[t].first = 0;
      dp[t].second++;
    }
    if (par[t] != -1) {
      dp[par[t]].second += dp[t].second;
      int len = dp[t].first;
      if (len > max1[par[t]]) swap(len, max1[par[t]]);
      if (len > max2[par[t]]) swap(len, max2[par[t]]);
    }
  }
  return dp[0].second;
}
void bin_solv(int l, int r, int l_val, int r_val) {
  if (l > r) {
    return;
  }
  if (l_val == r_val) {
    for (int i = l; i <= r; i++) {
      ans[i] = l_val;
    }
    return;
  }
  int m = (l + r) / 2;
  int m_val = calc(m);
  ans[m] = m_val;
  bin_solv(l, m - 1, m_val, r_val);
  bin_solv(m + 1, r, l_val, m_val);
}
void Solve() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    grp[u - 1].push_back(v - 1);
    grp[v - 1].push_back(u - 1);
  }
  dfs(0, -1);
  bin_solv(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << endl;
  }
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int TC = 1;
  while (TC--) Solve();
  return 0;
}
