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;
mt19937 rnd(228);
const int N = 1e5 + 7;
vector<int> g[N];
pair<int, int> dp[N];
int m1[N], m2[N];
int par[N];
vector<int> e;
void zhfs(int v, int pr) {
  par[v] = pr;
  for (int to : g[v]) {
    if (to != pr) {
      zhfs(to, v);
    }
  }
  e.push_back(v);
}
int n;
int f(int k) {
  for (int i = 0; i < n; i++) dp[i].first = dp[i].second = m1[i] = m2[i] = 0;
  for (int t : e) {
    if (m1[t] + m2[t] + 1 >= k) {
      dp[t].first = m1[t] + m2[t] + 1;
    } else {
      dp[t].first = m1[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 > m1[par[t]]) swap(len, m1[par[t]]);
      if (len > m2[par[t]]) swap(len, m2[par[t]]);
    }
  }
  return dp[0].second;
}
int ans[N];
void rec(int l, int r, int opt_l, int opt_r) {
  if (l > r) {
    return;
  }
  if (opt_l == opt_r) {
    for (int i = l; i <= r; i++) {
      ans[i] = opt_l;
    }
    return;
  }
  int m = (l + r) / 2;
  int opt_m = f(m);
  ans[m] = opt_m;
  rec(l, m - 1, opt_m, opt_r);
  rec(m + 1, r, opt_l, opt_m);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  n = 100000;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a = i, b = i + 1;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  zhfs(0, -1);
  rec(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << '\n';
  }
}
