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 maxN = 2e5;
vector<int> g[maxN];
int n;
vector<int> start[maxN];
int timer = 0;
int hg[maxN], ord[maxN];
void dfs(int v, int p) {
  timer++;
  ord[v] = timer;
  for (int to : start[v]) {
    if (to == p) continue;
    dfs(to, v);
    g[ord[v]].push_back(ord[to]);
  }
}
int calc_for(int v, int k) {
  if (k > n) return 0;
  int ans = 0;
  hg[v] = 0;
  int mx1 = 0, mx2 = 0;
  for (int to : g[v]) {
    ans += calc_for(to, k);
    if (hg[to] > mx1) {
      mx2 = mx1;
      mx1 = hg[to];
    } else
      mx2 = max(mx2, hg[to]);
  }
  if (mx1 + 1 + mx2 >= k) {
    hg[v] = 0;
    ans++;
  } else
    hg[v] = mx1 + 1;
  return ans;
}
int BLOCK = 1000;
int val[maxN];
int main() {
  srand(123 + 23123123 + 123123 - 123123);
  ios_base::sync_with_stdio(false);
  cin >> n;
  bool test = false;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    if (u == 36861 && i == 1) test = true;
    start[u].push_back(v);
    start[v].push_back(u);
  }
  dfs(1, -1);
  int sum = 0;
  for (int i = 1; i <= n; i++) sum += g[i].size();
  assert(sum == n - 1);
  memset(val, -1, sizeof val);
  BLOCK = min(BLOCK, n);
  for (int i = 1; i <= BLOCK; i++) {
    val[i] = calc_for(1, i);
  }
  vector<int> all;
  int st = BLOCK + 1;
  while (st <= n) {
    all.push_back(st);
    val[st] = calc_for(1, st);
    int l = st;
    int r = n;
    while (r - l > 0) {
      int mid = (r + l) / 2 + 1;
      if (calc_for(1, mid) == val[st])
        l = mid;
      else
        r = mid - 1;
    }
    l = st;
    for (; l <= r; l++) val[l] = val[st];
    st = r + 1;
  }
  for (int i = 1; i <= n; i++) cout << val[i] << '\n';
  return 0;
}
