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 nxt[N << 1], ver[N << 1], head[N], tot;
void addEdge(int u, int v) {
  nxt[++tot] = head[u];
  ver[tot] = v;
  head[u] = tot;
}
int tim, n;
int fa[N], dfn[N], ans[N], f[N];
void dfs(int u, int pre) {
  fa[u] = pre;
  for (int e = head[u]; e; e = nxt[e]) {
    int v = ver[e];
    if (v == pre) continue;
    dfs(v, u);
  }
  dfn[++tim] = u;
}
int solve(int k) {
  int res = 0;
  for (int i = 1; i <= n; ++i) f[i] = 1;
  for (int i = 1; i <= n; ++i) {
    int u = dfn[i];
    int fath = fa[u];
    if (fath && f[u] != -1 && f[fath] != -1) {
      if (f[u] + f[fath] >= k)
        res++, f[fath] = -1;
      else
        f[fath] = max(f[fath], f[u] + 1);
    }
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int u, v;
  int p = sqrt(n);
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &u, &v);
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0);
  ans[1] = n;
  for (int i = 2; i <= p; ++i) ans[i] = solve(i);
  for (int i = p + 1; i <= n; ++i) {
    int tmp = solve(i);
    int l = i, r = n;
    while (l < r) {
      int mid = l + r + 1 >> 1;
      if (solve(mid) == tmp)
        l = mid;
      else
        r = mid - 1;
    }
    for (; i <= r; ++i) ans[i] = tmp;
    --i;
  }
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
