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>
int n, u, v, sum[100005], len[100005], sz, rk[100005];
std::vector<int> ch[100005];
void dfs(int u, int f) {
  rk[++sz] = u;
  if (f) ch[u].erase(std::find(ch[u].begin(), ch[u].end(), f));
  for (int v : ch[u]) dfs(v, u);
}
int calc(int x) {
  std::memset(len, 0, sizeof(len));
  std::memset(sum, 0, sizeof(sum));
  for (int i = n; i >= 1; i--) {
    int u = rk[i];
    int fr = 0, sc = 0;
    for (int v : ch[u]) {
      if (len[v] > fr) {
        sc = fr;
        fr = len[v];
      } else if (len[v] > sc) {
        sc = len[v];
      }
      sum[u] += sum[v];
    }
    if (sc + fr + 1 >= x) {
      sum[u]++;
      len[u] = 0;
    } else {
      len[u] = fr + 1;
    }
  }
  return sum[1];
}
const int B = 1005;
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    ch[u].push_back(v);
    ch[v].push_back(u);
  }
  dfs(1, 0);
  int pre = 1;
  while (pre <= n) {
    int ans = calc(pre);
    printf("%d\n", ans);
    if (pre <= B) {
      pre++;
      continue;
    }
    for (int i = 17; i >= 0; i--) {
      int T = pre + (1 << i);
      if (T <= n && calc(T) == ans) {
        for (int j = pre + 1; j <= T; ++j) printf("%d\n", ans);
        pre = T;
      }
    }
    pre++;
  }
  return 0;
}
