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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
ifstream in("input.txt");
ofstream out("output.txt");
const int MAX_N = 1e5;
int n;
vector<int> g[MAX_N];
vector<int> ord;
int pr[MAX_N], D[MAX_N];
void dfs(int v, int p) {
  pr[v] = p;
  for (int to : g[v]) {
    if (to != p) dfs(to, v);
  }
  ord.push_back(v);
}
int calc(int k) {
  int res = 0;
  for (int i = n - 1; i >= 0; i--) {
    int v = ord[i];
    int max_d[2] = {0, 0};
    for (int to : g[v]) {
      if (to == pr[v]) continue;
      if (D[to] > max_d[0]) {
        max_d[1] = max_d[0];
        max_d[0] = D[to];
      } else if (D[to] > max_d[1])
        max_d[1] = D[to];
    }
    if (max_d[0] + max_d[1] + 1 >= k) {
      res++;
      D[v] = 0;
    } else
      D[v] = max_d[0] + 1;
  }
  return res;
}
void input() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  ;
  input();
  dfs(0, -1);
  reverse(ord.begin(), ord.end());
  int l = 1;
  while (l <= n) {
    int cur = calc(l);
    if (l <= 1000) {
      cout << cur << '\n';
      l++;
      continue;
    }
    int L = l, R = n + 1;
    while (R - L > 1) {
      int mid = (L + R) / 2;
      if (calc(mid) == cur)
        L = mid;
      else
        R = mid;
    }
    for (; l <= L; l++) cout << cur << '\n';
  }
}
