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>
const int maxN = 100005;
int N;
std::vector<int> G[maxN], G2[maxN];
int DFN[maxN], Index, Fa[maxN];
int Ans[maxN];
void DFS(int i) {
  DFN[i] = ++Index;
  for (auto to : G[i])
    if (to != Fa[i]) {
      Fa[to] = i;
      DFS(to);
      G2[DFN[i]].push_back(DFN[to]);
    }
}
int F[maxN];
int calc(int k) {
  int res = 0;
  for (int i = N; i; --i) {
    int mx = 0, smx = 0;
    for (auto to : G2[i]) {
      if (F[to] > mx)
        smx = mx, mx = F[to];
      else if (F[to] > smx)
        smx = F[to];
    }
    if (mx + smx + 1 >= k) {
      ++res;
      mx = -1;
    }
    F[i] = mx + 1;
  }
  return res;
}
void solve(int l, int r, int vL, int vR) {
  if (r - l <= 1) return;
  if (vL == vR) {
    for (int i = l + 1; i < r; ++i) Ans[i] = vL;
    return;
  }
  int m = (l + r) >> 1;
  Ans[m] = calc(m);
  solve(l, m, Ans[m], vR);
  solve(m, r, vL, Ans[m]);
}
int main() {
  scanf("%d", &N);
  for (int i = 1, u, v; i != N; ++i) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  DFS(1);
  solve(1, N, Ans[N] = calc(N), Ans[1] = N);
  for (int i = 1; i <= N; ++i) printf("%d\n", Ans[i]);
  return 0;
}
