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 n, tot, h[100005], ans[100005];
struct Edge {
  int to, nxt;
} d[100005 * 2];
void add(int x, int y) {
  d[++tot].to = y, d[tot].nxt = h[x];
  h[x] = tot;
}
int dp[100005], ret;
void dfs(int pos, int k, int f) {
  int mx = 0, sx = 0;
  dp[pos] = 0;
  for (int i = h[pos]; i; i = d[i].nxt) {
    int to = d[i].to;
    if (to == f) continue;
    dfs(to, k, pos);
    if (dp[to] > mx)
      sx = mx, mx = dp[to];
    else if (dp[to] > sx)
      sx = dp[to];
  }
  if (mx + sx + 1 >= k)
    ret++;
  else
    dp[pos] = mx + 1;
  return;
}
void CDQ(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = l + r >> 1;
  ret = 0;
  dfs(1, mid, 0);
  ans[mid] = ret;
  CDQ(l, mid - 1, ans[mid], R);
  CDQ(mid + 1, r, L, ans[mid]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  CDQ(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
