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 INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
vector<int> path[110000];
pair<int, int> dfs(int pos, int bef, int val) {
  pair<int, int> pa = make_pair(0, 0);
  pair<int, int> pb;
  int a = 0, b = 0;
  for (auto to : path[pos]) {
    if (to == bef) continue;
    pb = dfs(to, pos, val);
    pa.first += pb.first;
    if (a < pb.second) swap(a, pb.second);
    if (b < pb.second) swap(b, pb.second);
  }
  if (a + b + 1 >= val)
    pa.first++;
  else
    pa.second = a + 1;
  return pa;
}
const int B = 330;
int main() {
  int n, m;
  int i, j, k;
  int a, b, c;
  cin >> n;
  vector<int> vs(n);
  for (i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    path[a].push_back(b);
    path[b].push_back(a);
  }
  for (i = 0; i < B && i < n; i++) {
    vs[i] = dfs(0, -1, i + 1).first;
  }
  if (n <= B) {
    array_show(vs, vs.size(), '\n');
    return 0;
  }
  a = B;
  int z[3];
  vector<int> vz(B, n);
  for (i = B - 1; i >= 0; i--) {
    z[0] = a - 1, z[1] = vz[i];
    while (z[1] - z[0] > 1) {
      z[2] = (z[0] + z[1]) / 2;
      b = dfs(0, -1, z[2] + 1).first;
      if (b <= i) {
        z[1] = z[2];
        vz[b] = min(vz[b], z[2]);
      } else
        z[0] = z[2];
    }
    for (j = a; j < z[1]; j++) {
      vs[j] = i + 1;
    }
    for (j = 1; j < i; j++) {
      vz[j] = min(vz[j], vz[j - 1]);
    }
    a = z[1];
  }
  array_show(vs, vs.size(), '\n');
}
