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;
int const nmax = 100000;
vector<int> g[1 + nmax];
void build(int node, int parent) {
  for (int h = 0; h < g[node].size(); h++) {
    int to = g[node][h];
    if (to == parent) g[node].erase(g[node].begin() + h);
  }
  for (int h = 0; h < g[node].size(); h++) build(g[node][h], node);
}
pair<int, int> dp[1 + nmax];
void dfs(int node, int parent, int k) {
  int smax1 = 0, smax2 = 0, result = 0;
  for (int h = 0; h < g[node].size(); h++) {
    int to = g[node][h];
    dfs(to, node, k);
    result += dp[to].first;
    if (smax1 < dp[to].second) {
      smax2 = smax1;
      smax1 = dp[to].second;
    } else if (smax2 < dp[to].second) {
      smax2 = dp[to].second;
    }
  }
  smax1++;
  if (k <= smax1 + smax2)
    dp[node] = {result + 1, 0};
  else
    dp[node] = {result, smax1};
}
int sol[1 + nmax];
int eval(int k) {
  if (sol[k] == -1) {
    dfs(1, 0, k);
    sol[k] = dp[1].first;
  }
  return sol[k];
}
int binarysearch(int from, int to, int target) {
  if (from < to) {
    int mid = (from + to + 1) / 2;
    if (target <= eval(mid))
      return binarysearch(mid, to, target);
    else
      return binarysearch(from, mid - 1, target);
  } else
    return from;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) sol[i] = -1;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(1, 0);
  int ptr = 1;
  while (ptr <= n) {
    int val = eval(ptr);
    int pos = binarysearch(1, n, val);
    while (ptr <= pos) sol[ptr++] = val;
  }
  for (int i = 1; i <= n; i++) cout << sol[i] << '\n';
  return 0;
}
