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 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, m, k;
vector<pair<int, int>> dp;
vector<vector<int>> G;
map<int, int> Memo;
vector<int> Ans;
void f(int v, int k, int p = -1) {
  int mx1 = 0, mx2 = 0;
  for (int y : G[v]) {
    if (y == p) continue;
    f(y, k, v);
    if (dp[y].first > mx1) swap(mx1, dp[y].first);
    if (dp[y].first > mx2) swap(mx2, dp[y].first);
    dp[v].second += dp[y].second;
  }
  if (mx1 + mx2 + 1 >= k) {
    dp[v].second++;
  } else {
    dp[v].first = mx1 + 1;
  }
}
void get(int l, int r, int cl, int cr) {
  if (l > r) return;
  if (cl == cr) {
    for (int i = l - 1; i < r; i++) {
      Ans[i] = cl;
    }
  } else {
    int m = (l + r) >> 1;
    dp.assign(n, {0, 0});
    f(0, m);
    cerr << m << " : " << dp[0].second << "\n";
    Ans[m - 1] = dp[0].second;
    get(l, m - 1, Ans[m - 1], cr);
    get(m + 1, r, cl, Ans[m - 1]);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t zzz = clock();
  cin >> n;
  G.resize(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  Ans.resize(n);
  get(1, n, 0, n);
  for (auto x : Ans) cout << x << "\n";
}
