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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, now, cnt;
int f[200010], ans[200010];
struct edge {
  int to, nxt;
} e[200010];
int head[200010], edge_cnt;
void add(int from, int to) {
  e[++edge_cnt] = (edge){to, head[from]};
  head[from] = edge_cnt;
}
void dfs(int x, int fa) {
  f[x] = 0;
  int m1 = 0, m2 = 0;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs(y, x);
    if (f[y] > m1)
      m2 = m1, m1 = f[y];
    else if (f[y] > m2)
      m2 = f[y];
  }
  if (m1 + m2 + 1 >= now)
    cnt++;
  else
    f[x] = m1 + 1;
}
void solve(int L, int R, int l, int r) {
  if (L > R || l > r) return;
  if (l == r) {
    for (int i = L; i <= R; ++i) ans[i] = l;
    return;
  }
  int mid = (L + R) >> 1;
  now = mid, cnt = 0, dfs(1, 0), ans[mid] = cnt;
  solve(L, mid - 1, cnt, r), solve(mid + 1, R, l, cnt);
}
int main() {
  read(n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    read(x), read(y);
    add(x, y), add(y, x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
