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 s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
  return f ? s : -s;
}
const int N = 1e5 + 10;
int n;
int tot, head[N], nxt[N << 1], ver[N << 1];
void add(int x, int y) {
  nxt[++tot] = head[x];
  ver[tot] = y;
  head[x] = tot;
}
int f[N], cnt;
void dfs(int x, int _fa, int k) {
  f[x] = 0;
  int fMax = 0, sMax = 0;
  for (register int i = head[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == _fa) continue;
    dfs(y, x, k);
    if (f[y] > fMax)
      sMax = fMax, fMax = f[y];
    else if (f[y] > sMax)
      sMax = f[y];
  }
  if (fMax + sMax + 1 >= k)
    f[x] = 0, ++cnt;
  else
    f[x] = fMax + 1;
}
int ans[N];
void sol(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (register int i = l; i <= r; ++i) ans[i] = L;
    return;
  }
  int mid = l + r >> 1;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  sol(l, mid - 1, cnt, R);
  sol(mid + 1, r, L, cnt);
}
int main() {
  n = read();
  for (register int i = 1; i < n; ++i) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  sol(1, n, 0, n);
  for (register int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
