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 << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
inline void write(int x) {
  if (!x) return putchar('0'), void();
  int q[105], t = 0;
  while (x) q[++t] = x % 10, x /= 10;
  while (t) putchar(q[t--] ^ '0');
}
const int N = 1e5 + 5;
struct edge {
  int y, next;
} e[N << 1];
int cnt, first[N];
inline void add(int x, int y) {
  e[++cnt] = {y, first[x]};
  first[x] = cnt;
}
int f[N], g[N], fat[N], ord[N], tot;
int n, ans[N];
void dfs(int x, int fa) {
  fat[x] = fa;
  for (int i = first[x]; i; i = e[i].next) {
    int y = e[i].y;
    if (y == fa) continue;
    dfs(y, x);
  }
  ord[++tot] = x;
}
inline int get(int k) {
  int res = 0;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) {
    int x = ord[i], fa = fat[x];
    if (fa && f[x] > 0 && f[fa] > 0)
      if (f[x] + f[fa] >= k)
        res++, f[fa] = -1;
      else
        f[fa] = max(f[fa], f[x] + 1);
  }
  return res;
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  int B = sqrt(n * log2(n));
  dfs(1, 0);
  ans[1] = n;
  for (int i = 2; i <= B; i++) ans[i] = get(i);
  for (int l = B + 1, r = n; l <= n; l = r + 1, r = n) {
    const int x = get(l);
    int L = l;
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (get(mid) == x)
        l = mid;
      else
        r = mid - 1;
    }
    for (int i = L; i <= l; i++) ans[i] = x;
  }
  for (int i = 1; i <= n; i++) write(ans[i]), puts("");
  return 0;
}
