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 maxn = 100010;
int n, cnt, fa[maxn], pos[maxn], ans[maxn], mx[maxn], head[maxn];
struct edge {
  int to, nxt;
} e[maxn << 1];
void add(int u, int v) { e[++cnt] = (edge){v, head[u]}, head[u] = cnt; }
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    add(u, v), add(v, u);
  }
  int tim = 0;
  function<void(int, int)> init = [&](int v, int fa) {
    pos[++tim] = v, ::fa[v] = fa;
    for (int i = head[v]; i; i = e[i].nxt)
      if (e[i].to ^ fa) {
        init(e[i].to, v);
      }
  };
  init(1, 0);
  memset(ans, -1, sizeof(ans));
  auto solve = [&](int k) {
    if (~ans[k]) return ans[k];
    if (k == 1) return ans[k] = n;
    ans[k] = 0;
    for (int _ = n; _; _--) {
      int v = pos[_], fir = 0, sec = 0, fa = ::fa[v];
      for (int i = head[v]; i; i = e[i].nxt)
        if (e[i].to ^ fa) {
          if (mx[e[i].to] > fir)
            sec = fir, fir = mx[e[i].to];
          else
            sec = max(sec, mx[e[i].to]);
        }
      if (fir + sec >= k - 1)
        ans[k]++, mx[v] = 0;
      else
        mx[v] = fir + 1;
    }
    return ans[k];
  };
  function<void(int, int)> divide = [&](int l, int r) {
    if (solve(l) == solve(r)) {
      fill(ans + l, ans + r + 1, solve(l));
      return;
    }
    int mid = (l + r) >> 1;
    divide(l, mid), divide(mid + 1, r);
  };
  divide(1, n);
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
