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 = 100005;
struct Edg {
  int lst, ed;
} e[N << 1];
int ne = 0, nte[N], f[N];
int tot = 0, dfn[N], fa[N];
int Mx[N], sMx[N];
int rd() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}
void NewEdg(int u, int v) {
  ne++;
  e[ne].lst = nte[u];
  e[ne].ed = v;
  nte[u] = ne;
}
void dfs(int x) {
  dfn[++tot] = x;
  for (int i = nte[x]; i; i = e[i].lst) {
    if (e[i].ed == fa[x]) continue;
    fa[e[i].ed] = x;
    dfs(e[i].ed);
  }
}
int Sol(int w) {
  int res = 0;
  for (int i = 1; i <= tot; i++) Mx[i] = sMx[i] = f[i] = 0;
  for (int i = tot; i >= 1; i--) {
    int x = dfn[i];
    if (sMx[x] + Mx[x] + 1 >= w)
      res++, f[x] = 0;
    else
      f[x] = Mx[x] + 1;
    if (x == 1) break;
    if (f[x] > Mx[fa[x]])
      sMx[fa[x]] = Mx[fa[x]], Mx[fa[x]] = f[x];
    else if (f[x] > sMx[fa[x]])
      sMx[fa[x]] = f[x];
  }
  return res;
}
int main() {
  int n = rd();
  for (int i = 1; i <= n - 1; i++) {
    int u = rd(), v = rd();
    NewEdg(u, v), NewEdg(v, u);
  }
  dfs(1);
  int tn = sqrt(n);
  for (int i = 1; i <= tn; i++) printf("%d\n", Sol(i));
  for (int i = tn + 1; i <= n; i++) {
    int ans = Sol(i);
    int l = i, r = n + 1;
    while (l + 1 < r) {
      int mid = (l + r) >> 1;
      if (Sol(mid) == ans)
        l = mid;
      else
        r = mid;
    }
    for (int j = i; j <= l; j++) printf("%d\n", ans);
    i = l;
  }
  return 0;
}
