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 + 5;
vector<int> g[N];
int ret = 0;
int calc(int u, int p, int k) {
  pair<int, int> mx = {0, 0};
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    mx.first = max(mx.first, calc(v, u, k) + 1);
    if (mx.first > mx.second) swap(mx.first, mx.second);
  }
  mx.second = max(mx.second, 1);
  int temp;
  if (!mx.first) {
    temp = mx.second;
    if (temp >= k) {
      ret++;
      temp = 0;
    }
  } else {
    temp = mx.second + mx.first - 1;
    if (temp >= k) {
      ret++;
      temp = 0;
    } else
      temp = mx.second;
  }
  return temp;
}
int res[N];
void bs(int l, int r, int lb, int ub) {
  if (l > r) return;
  if (lb == ub) {
    for (int i = l; i <= r; i++) {
      res[i] = ub;
    }
    return;
  }
  int mid = (l + r) / 2;
  ret = 0;
  calc(1, -1, mid);
  res[mid] = ret;
  bs(l, mid - 1, lb, ret);
  bs(mid + 1, r, ret, ub);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    swap(a, b);
    g[a].push_back(b);
  }
  int a, b;
  ret = 0;
  calc(1, -1, 1);
  a = ret;
  ret = 0;
  calc(1, -1, n);
  b = ret;
  bs(1, n, a, b);
  for (int i = 1; i <= n; i++) {
    printf("%d ", res[i]);
  }
  return 0;
}
