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 = 1e5 + 5;
int n, dp[MAXN], ans[MAXN];
vector<int> g[MAXN], c[MAXN];
vector<int> ord;
void dfs(int v, int f = -1) {
  for (int u : g[v]) {
    if (u == f) continue;
    c[v].push_back(u);
    dfs(u, v);
  }
  ord.push_back(v);
}
int comp(int k) {
  int tot = 0;
  for (int i = 0; i < n; ++i) {
    int v1 = 0, v2 = 0;
    int v = ord[i];
    for (int u : c[v]) {
      if (dp[u] > v1) {
        v2 = v1;
        v1 = dp[u];
      } else if (dp[u] > v2) {
        v2 = dp[u];
      }
    }
    if (1 + v1 + v2 >= k) {
      dp[v] = 0;
      ++tot;
    } else {
      dp[v] = 1 + v1;
    }
  }
  ans[k] = tot;
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 1; i <= min(n, 1000); ++i) ans[i] = comp(i);
  int x = 1001;
  while (true) {
    if (x > n) break;
    int y = comp(x);
    int lo = x, hi = n;
    while (lo != hi) {
      int m = (lo + hi + 1) / 2;
      if (comp(m) == y)
        lo = m;
      else
        hi = m - 1;
    }
    for (int i = x; i <= lo; ++i) ans[i] = y;
    x = lo + 1;
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << endl;
}
