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 N = 1e5 + 5;
int read() {
  int x, f = 1;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (f = -f);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 3) + (x << 1) + (ch ^ 48))
    ;
  return x * f;
}
struct Edge {
  int to;
  Edge *nxt;
  Edge(int to, Edge *nxt) : to(to), nxt(nxt) {}
} * head[N];
void add(int x, int y) { head[x] = new Edge(y, head[x]); }
int n;
namespace solve1 {
int cnt, f[N];
void dfs(int x, int fa, int k) {
  int mx1 = 0, mx2 = 0;
  for (Edge *i = head[x]; i; i = i->nxt) {
    if (i->to == fa) continue;
    dfs(i->to, x, k);
    if (f[i->to] > mx1)
      mx2 = mx1, mx1 = f[i->to];
    else if (f[i->to] > mx2)
      mx2 = f[i->to];
  }
  if (mx1 + mx2 + 1 >= k)
    f[x] = 0, cnt++;
  else
    f[x] = mx1 + 1;
}
void work() {
  int i;
  for (i = 1; i <= n; ++i) {
    cnt = 0;
    dfs(1, 0, i);
    printf("%d\n", cnt);
    if (cnt == 0) {
      ++i;
      break;
    }
  }
  for (; i <= n; ++i) puts("0");
}
}  // namespace solve1
namespace solve2 {
int cnt, f[N], ans[N];
void dfs(int x, int fa, int k) {
  int mx1 = 0, mx2 = 0;
  for (Edge *i = head[x]; i; i = i->nxt) {
    if (i->to == fa) continue;
    dfs(i->to, x, k);
    if (f[i->to] > mx1)
      mx2 = mx1, mx1 = f[i->to];
    else if (f[i->to] > mx2)
      mx2 = f[i->to];
  }
  if (mx1 + mx2 + 1 >= k)
    f[x] = 0, cnt++;
  else
    f[x] = mx1 + 1;
}
void solve(int l, int r, int x, int y) {
  if (x == y) {
    for (int i = l; i <= r; ++i) ans[i] = x;
    return;
  }
  int mid = (l + r) >> 1;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  if (l <= mid - 1 && cnt <= y) solve(l, mid - 1, cnt, y);
  if (mid + 1 <= r && x <= cnt) solve(mid + 1, r, x, cnt);
}
void work() {
  solve(1, n, 0, n);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
}
}  // namespace solve2
int main() {
  n = read();
  for (int i = 1, x, y; i < n; ++i)
    x = read(), y = read(), add(x, y), add(y, x);
  solve2::work();
  fclose(stdin);
  fclose(stdout);
  return 0;
}
