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 GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
using namespace std;
int a, b, c, d, e, f, p = 1000000007, base = 157, mi = 1e9, id = 0, q, pos;
int ind[300000], ans[300000], si[300000], si2[300000];
pair<int, int> reb_price[300000];
string s, s2;
vector<int> vec[300000];
vector<pair<int, int>> rebra[300000];
vector<int> order;
void dfs() {
  for (auto v : order) {
    int ma = 0, ma2 = 0;
    si[v] = 1;
    for (auto i : rebra[v]) {
      if (i.second == 0) {
        continue;
      }
      b += (si[i.first] + i.second - 1) / id;
      si[i.first] = (si[i.first] + i.second - 1) % id;
      if (si[i.first] > ma) {
        ma2 = ma;
        ma = si[i.first];
      } else if (si[i.first] > ma2) {
        ma2 = si[i.first];
      }
    }
    si[v] = ma + 1;
    if (si[v] + ma2 >= id) {
      si[v] = 0;
      b++;
    }
  }
}
pair<int, int> dfs2(int v, int pred, int pred2) {
  if (rebra[v].size() == 1 && v != 1) {
    rebra[v][0] = {pred2, 0};
    order.push_back(v);
    return {v, 1};
  }
  if (rebra[v].size() == 2 && v != 1) {
    if (rebra[v][0].first == pred) {
      auto pa = dfs2(rebra[v][1].first, v, pred2);
      return {pa.first, pa.second + 1};
    } else {
      auto pa = dfs2(rebra[v][0].first, v, pred2);
      return {pa.first, pa.second + 1};
    }
  }
  int j = 0;
  for (auto i : rebra[v]) {
    if (i.first == pred) {
      rebra[v][j] = {pred2, 0};
      j++;
      continue;
    }
    rebra[v][j] = dfs2(i.first, v, v);
    j++;
  }
  order.push_back(v);
  return {v, 1};
}
signed main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
  cin >> a;
  for (int i = 0; i < a - 1; i++) {
    cin >> b >> c;
    rebra[b].push_back({c, 1});
    rebra[c].push_back({b, 1});
  }
  dfs2(1, -1, -1);
  for (int i = 0; i < a; i = i) {
    if (i == 0 || (b > 100 && i < 500)) {
      b = 0;
      id = i + 1;
      dfs();
      cout << b << '\n';
      ;
      i++;
    } else {
      b = 0;
      id = i + 1;
      dfs();
      c = b;
      int l = i + 1, r = a / max(b, 1) + 1;
      b = 0;
      while (r - l > 1) {
        int res = (r + l) / 2;
        b = 0;
        id = res;
        dfs();
        if (b == c) {
          l = res;
        } else {
          r = res;
        }
      }
      for (int j = i; j < l; j++) {
        cout << c << '\n';
        ;
      }
      i = l;
    }
  }
  cin >> a;
  return 0;
}
