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 = 100005;
pair<int, int> dp[N];
int mark[N], pass;
vector<int> g[N], h[N];
int id[N], ne, n;
void prec(int u, int par = -1) {
  for (int v : h[u])
    if (v != par) {
      prec(v, u);
    }
  id[u] = ++ne;
  for (int v : h[u])
    if (v != par) g[id[u]].push_back(id[v]);
}
int k;
int solve() {
  for (int u = 1; u <= n; u++) {
    auto &ans = dp[u];
    ans.first = 0, ans.second = 1;
    int a = 0, b = 0;
    for (int v : g[u]) {
      auto &tmp = dp[v];
      ans.first += tmp.first;
      a = max(a, tmp.second);
      if (a > b) swap(a, b);
    }
    if (a + b + 1 >= k)
      ans.first++, ans.second = 0;
    else
      ans.second = b + 1;
  }
  return dp[n].first;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d %d", &a, &b);
    h[a].push_back(b);
    h[b].push_back(a);
  }
  prec(1);
  printf("%d\n", n);
  for (int i = 2; i <= n; i++) {
    k = i;
    int tmp = solve(), L = i + 1, R = n, ans = i;
    while (L <= R) {
      k = (L + R) / 2;
      if (solve() == tmp)
        L = k + 1, ans = k;
      else
        R = k - 1;
    }
    for (int j = i; j <= ans; j++) printf("%d\n", tmp);
    i = ans;
  }
}
