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, ans[100010], f[100010], head[500010], fr[500010], nxt[500010], dis,
    cnt = 0;
void add(int x, int y) {
  head[++cnt] = y;
  nxt[cnt] = fr[x];
  fr[x] = cnt;
}
void dfs(int x, int fa, int check) {
  f[x] = 0;
  int flag = 0;
  for (int i = fr[x]; i; i = nxt[i]) {
    int to = head[i];
    if (to == fa) continue;
    dfs(to, x, check);
    if (f[x] + f[to] >= check - 1) flag = 1;
    f[x] = max(f[x], f[to]);
  }
  if (flag || check == 1) {
    dis++;
    f[x] = 0;
  } else
    f[x]++;
}
void merge(int l, int r, int ansl, int ansr) {
  if (l > r) return;
  if (ansl == ansr) {
    for (int i = l; i <= r; i++) ans[i] = ansl;
    return;
  }
  int mid = (l + r) >> 1;
  dis = 0;
  dfs(1, 0, mid);
  ans[mid] = dis;
  merge(l, mid - 1, ans[mid], ansr);
  merge(mid + 1, r, ansl, ans[mid]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  merge(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
}
