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], f[N];
int la2[N], ne2[N], tt[N];
void get(int o) {
  int ts = 0, g;
  for (int i = n, _e = 1; i >= _e; i--) {
    x = d[i];
    g = 0;
    for (int i = la2[x]; i && ~g; i = ne2[i]) {
      if (g + f[tt[i]] + 1 >= o)
        ++ts, g = -1;
      else if (g < f[tt[i]])
        g = f[tt[i]];
    }
    f[x] = g + 1;
  }
  an[o] = ts;
}
void calc(int l, int r, int vl, int vr) {
  if (r - l <= 1) return;
  if (vl == vr) {
    for (int i = l, _e = r; i <= _e; i++) an[i] = vl;
    return;
  }
  int mi = l + r >> 1;
  get(mi);
  int vm = an[mi];
  calc(l, mi, vl, vm);
  calc(mi, r, vm, vr);
}
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);
  js = 0;
  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],
        (ne2[++js] = la2[x], la2[x] = js, tt[js] = to[i]);
  get(n);
  an[1] = n;
  calc(1, n, n, an[n]);
  for (int i = 1, _e = n; i <= _e; i++) printf("%d\n", an[i]);
}
