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;
int dp[(int)(1e5 + 10)], dp2[(int)(1e5 + 10)], pre[(int)(1e5 + 10)];
int i, j, k, n, m, x, y, z, ans[(int)(1e5 + 10)];
vector<int> arr[(int)(1e5 + 10)], v, arr2[(int)(1e5 + 10)];
void f(int x, int root) {
  for (int j = 0; j < n; j++) {
    int x = v[j];
    dp2[x] = dp[x] = 0;
    int mx1 = 0, mx2 = 0;
    for (int i = 0; i < arr2[x].size(); i++) {
      int y = arr2[x][i];
      if (mx1 < dp2[y]) {
        mx2 = mx1;
        mx1 = dp2[y];
      } else if (mx2 < dp2[y])
        mx2 = dp2[y];
      dp[x] += dp[y];
    }
    if (mx1 + mx2 + 1 >= k)
      dp[x]++;
    else
      dp2[x] = mx1 + 1;
  }
  return;
}
int f2(int t) {
  int bas = 1;
  int son = n + 1;
  while (bas < son) {
    k = ((bas + son) >> 1);
    f(1, 0);
    int x = dp[1];
    if (x >= t)
      bas = ((bas + son) >> 1) + 1;
    else
      son = ((bas + son) >> 1);
  }
  return bas;
}
int main() {
  cin >> n;
  for (i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    arr[x].push_back(y);
    arr[y].push_back(x);
  }
  queue<int> xx;
  xx.push(1);
  while (xx.size()) {
    int x = xx.front();
    v.push_back(x);
    xx.pop();
    for (int i = 0; i < arr[x].size(); i++) {
      if (pre[x] != arr[x][i]) {
        arr2[x].push_back(arr[x][i]);
        xx.push(arr[x][i]);
        pre[arr[x][i]] = x;
      }
    }
  }
  reverse(v.begin(), v.end());
  for (i = 1; i <= min(1500, n); i++) {
    k = i;
    f(1, 0);
    ans[i] = dp[1];
    if (!ans[i]) break;
  }
  int last = i - 1;
  int t = ans[i - 1];
  while (t and last <= n) {
    int r = f2(t);
    for (i = last; i < r; i++) ans[i] = t;
    last = r;
    t--;
  }
  for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
}
