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 mod = 1e9 + 7;
const int N = 700012;
const int inf = 2e9;
int n, u, v;
vector<int> g[N];
int k, ans;
int best[N];
int len[N];
int st;
vector<int> order;
int par[N];
int was[N], dp[N];
inline int dfs(int st) {
  if (was[st]) return dp[st];
  was[st] = 1;
  int ans = 0;
  for (int v : order) {
    int mx1 = 0, mx2 = -1e9;
    for (int to : g[v]) {
      if (to == par[v]) continue;
      if (len[to] >= mx1)
        mx2 = mx1, mx1 = len[to];
      else if (len[to] >= mx2)
        mx2 = len[to];
    }
    if (mx1 + mx2 + 1 >= st) {
      len[v] = 0;
      ++ans;
    } else {
      len[v] = mx1 + 1;
    }
  }
  return dp[st] = ans;
}
int lvl[N];
inline void calc(int v, int p = 0) {
  lvl[v] = lvl[p] + 1;
  par[v] = p;
  for (int to : g[v]) {
    if (to == p) continue;
    calc(to, v);
  }
  order.push_back(v);
}
int res[N];
inline void solve(int l, int r, int optl, int optr) {
  if (l + 1 >= r) return;
  if (optl == optr) {
    for (int i = l; i <= r; ++i) res[i] = optl;
    return;
  }
  int m = l + r >> 1;
  res[m] = dfs(m);
  solve(l, m, optl, res[m]);
  solve(m, r, res[m], optr);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  calc(1);
  was[1] = 1;
  dp[1] = n;
  res[n] = dfs(n);
  res[1] = dfs(1);
  solve(1, n, res[1], res[n]);
  for (int i = 1; i <= n; ++i) cout << res[i] << '\n';
  return 0;
}
