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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
inline int max(int a, int b) { return a > b ? a : b; }
const int N = 2e5 + 10;
int n, t = 0, linkk[N], a[N], cnt = 0, p[N], f[N], d[N], q, ans[N];
struct node {
  int y, next;
} e[N * 2];
void insert(int x, int y) {
  e[++t].y = y;
  e[t].next = linkk[x];
  linkk[x] = t;
}
void dfs(int x, int father) {
  for (int i = linkk[x]; i; i = e[i].next) {
    int y = e[i].y;
    if (y == father) continue;
    dfs(y, x);
  }
  d[++cnt] = x;
  p[x] = father;
}
int solve(int k) {
  int r = 0;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) {
    int x = d[i];
    if (p[x] && (~f[p[x]]) && (~f[x]))
      if (f[x] + f[p[x]] >= k)
        r++, f[p[x]] = -1;
      else
        f[p[x]] = max(f[p[x]], f[x] + 1);
  }
  return r;
}
int main() {
  n = read();
  q = sqrt(n * log(n) / log(2));
  for (int i = 1; i < n; i++) {
    int x, y;
    x = read();
    y = read();
    insert(x, y);
    insert(y, x);
  }
  dfs(1, 0);
  ans[1] = n;
  for (int i = 2; i <= q; i++) ans[i] = solve(i);
  int l, r;
  for (int i = q + 1; i <= n; i = l + 1) {
    l = i, r = n;
    int t = solve(i), mid;
    for (; l < r; solve(mid) ^ t ? r = mid - 1 : l = mid) mid = l + r + 1 >> 1;
    for (int j = i; j <= l; j++) ans[j] = t;
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
