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 = 100005;
int n, h, firs[maxn], ans[maxn], dep[maxn], k, x, y;
int len[maxn], f[maxn], t[maxn], sta[maxn], tp;
struct edge {
  int too, nex;
} g[maxn << 1];
vector<int> v[maxn];
void add(int x, int y) {
  g[++h].nex = firs[x];
  firs[x] = h;
  g[h].too = y;
}
void dfs(int x) {
  int j;
  for (register int i = firs[x]; i; i = g[i].nex) {
    j = g[i].too;
    if (dep[j]) continue;
    v[x].push_back(j);
    dep[j] = dep[x] + 1;
    dfs(j);
  }
  sta[++tp] = x;
}
void dp() {
  for (register int t = 1; t <= tp; ++t) {
    int x = sta[t];
    int j, m1 = 0, m2 = 0;
    f[x] = 0;
    int s = v[x].size();
    for (register int i = 0; i < s; i++) {
      j = v[x][i];
      if (len[j] > m1)
        m2 = m1, m1 = len[j];
      else
        m2 = max(m2, len[j]);
      f[x] += f[j];
    }
    if (m1 + m2 + 1 < k)
      len[x] = m1 + 1;
    else
      len[x] = 0, f[x]++;
  }
}
int getans(int x) {
  if (t[x] != -1) return t[x];
  if (x == 1) return n;
  k = x;
  dp();
  t[x] = f[1];
  return f[1];
}
inline void solve(int s, int t, int l, int r) {
  if (s > t || l > r) return;
  if (l == r) {
    for (register int i = s; i <= t; ++i) ans[i] = l;
    return;
  }
  int mid = (s + t) >> 1;
  ans[mid] = getans(mid);
  solve(s, mid - 1, ans[mid], r);
  solve(mid + 1, t, l, ans[mid]);
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dep[1] = 1;
  dfs(1);
  for (register int i = 1; i <= n; ++i) t[i] = -1;
  solve(1, n, 0, n);
  for (register int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
