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;
int n, x, y, la[N], ne[N * 2], to[N * 2], js, an[N], d[N], fa[N];
int f[N], fx[N], wz;
void get(int o) {
  int tt = 0;
  for (int i = 1, _e = n; i <= _e; i++) f[i] = fx[i] = 0;
  for (int i = n, _e = 1; i >= _e; i--) {
    x = d[i];
    y = fa[x];
    if (++f[x] + fx[x] >= o) tt++, f[x] = 0;
    if (f[x] > f[y])
      fx[y] = f[y], f[y] = f[x];
    else
      fx[y] = (fx[y] > f[x] ? fx[y] : f[x]);
  }
  an[o] = tt;
}
int main() {
  scanf("%d", &n);
  for (int i = 2, _e = n; i <= _e; i++)
    scanf("%d%d", &x, &y), (ne[++js] = la[x], la[x] = js, to[js] = y),
        (ne[++js] = la[y], la[y] = js, to[js] = x);
  for (int t = 0, w = d[1] = 1; x = d[t + 1], t++ < w;)
    for (int i = la[x]; i; i = ne[i])
      if (to[i] != fa[x]) fa[to[i]] = x, d[++w] = to[i];
  for (int i = 1, _e = n; i <= _e; i++) an[i] = -1;
  an[1] = n;
  for (int i = 2, _e = n; i <= _e; i++) {
    if (an[i] == -1) get(i);
    wz = i;
    for (int l = i + 1, r = n, mi; mi = l + r >> 1, l <= r;) {
      if (an[mi] == -1) get(mi);
      if (an[i] == an[mi])
        wz = mi, l = mi + 1;
      else
        r = mi - 1;
    }
    for (int j = i + 1, _e = wz; j <= _e; j++) an[j] = an[i];
    i = wz;
  }
  for (int i = 1, _e = n; i <= _e; i++) printf("%d\n", an[i]);
}
