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 + 10;
int n, tot, len;
int fir[N], f[N], leaf[N], ans[N];
struct edge {
  int to;
  int nxt;
} e[N << 1];
inline void add(int x, int y) {
  e[++tot].to = y;
  e[tot].nxt = fir[x];
  fir[x] = tot;
  e[++tot].to = x;
  e[tot].nxt = fir[y];
  fir[y] = tot;
}
inline void dfs(int p, int fa) {
  int maxn = 1;
  leaf[p] = 1;
  for (int i = fir[p]; i; i = e[i].nxt) {
    int q = e[i].to;
    if (q == fa) continue;
    dfs(q, p), f[p] += f[q];
    maxn = max(maxn, leaf[p] + leaf[q]);
    leaf[p] = max(leaf[p], leaf[q] + 1);
  }
  if (maxn >= len) f[p]++, leaf[p] = 0;
}
inline void solve(int l, int r, int x, int y) {
  if (x == y) {
    for (int i = l; i <= r; i++) ans[i] = x;
    return;
  }
  int mid = (l + r) >> 1;
  for (int i = 1; i <= n; i++) f[i] = leaf[i] = 0;
  len = mid, dfs(1, 0), ans[mid] = f[1];
  if (l < mid) solve(l, mid - 1, ans[mid], y);
  if (r > mid) solve(mid + 1, r, x, ans[mid]);
}
int main() {
  int a, b;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d", &a, &b), add(a, b);
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
