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 N = 1e5 + 5;
vector<int> edge[N];
int ans[N];
int dfs(int u, int pre, int k) {
  int max1 = 0, max2 = 0;
  for (auto v : edge[u]) {
    if (v == pre) continue;
    int maxLen = dfs(v, u, k);
    ans[k] += maxLen / k;
    maxLen = maxLen % k;
    if (max1 == 0) {
      max1 = maxLen;
      continue;
    }
    if (max1 <= maxLen) {
      max2 = max1;
      max1 = maxLen;
      continue;
    }
    if (max2 < maxLen) {
      max2 = maxLen;
    }
  }
  if (max1 + max2 + 1 >= k) {
    ++ans[k];
    return 0;
  }
  return max1 + 1;
}
void DNC(int l, int r) {
  if (l + 1 >= r) {
    return;
  }
  if (ans[l] == ans[r]) {
    for (int i = l + 1; i <= r - 1; ++i) ans[i] = ans[l];
    return;
  }
  int mid = (l + r) >> 1;
  dfs(1, -1, mid);
  DNC(l, mid);
  DNC(mid, r);
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(1, -1, 1);
  dfs(1, -1, n);
  DNC(1, n);
  for (int i = 1; i <= n; ++i) cout << ans[i] << "\n";
}
