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 + 1;
int n;
vector<int> g[N];
pair<int, int> dfs(int cur, int p, const int &k) {
  int v[2] = {0, 0};
  int complete = 0;
  for (auto nxt : g[cur])
    if (nxt != p) {
      auto ret = dfs(nxt, cur, k);
      complete += ret.first;
      int incomplete = ret.second;
      for (int i = 0; i < 2; ++i)
        if (incomplete > v[i]) swap(v[i], incomplete);
    }
  if (v[0] + v[1] + 1 >= k)
    return make_pair(complete + 1, 0);
  else
    return make_pair(complete, v[0] + 1);
}
void solve(int kl, int kr, int vl, int vr) {
  if (vl == vr) {
    for (int i = kl; i <= kr; ++i) printf("%d\n", vl);
    return;
  }
  int km = (kl + kr) >> 1;
  int vm = dfs(1, 1, km).first;
  if (kl < km) solve(kl, km - 1, vm, vr);
  printf("%d\n", vm);
  if (km < kr) solve(km + 1, kr, vl, vm);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int k, ans;
  for (k = 1; k * k < n; ++k) printf("%d\n", ans = dfs(1, 1, k).first);
  solve(k, n, 0, ans);
}
