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 + 10;
vector<int> edge[N];
int n, x, y, dfn[N], f[N], cnt, ans[N], p = 1, tree[N][5];
void dfs(int x, int fa) {
  f[x] = fa;
  dfn[++cnt] = x;
  for (int i = 0; i < edge[x].size(); i++) {
    int v = edge[x][i];
    if (v != fa) dfs(v, x);
  }
}
int solve(int x) {
  if (~ans[x]) return ans[x];
  int res = 0;
  for (int i = 0; i <= n; i++) tree[i][0] = tree[i][1] = 0;
  for (int i = n; i >= 1; i--) {
    int u = dfn[i];
    for (int j = 0; j < edge[u].size(); j++) {
      int v = edge[u][j];
      if (v != f[u]) {
        int s = tree[v][0];
        if (tree[u][0] < s) {
          tree[u][1] = tree[u][0];
          tree[u][0] = s;
        } else if (s > tree[u][1]) {
          tree[u][1] = s;
        }
      }
    }
    tree[u][0]++;
    if (tree[u][0] + tree[u][1] >= x) res++, tree[u][0] = 0;
  }
  ans[x] = res;
  return res;
}
void Solve(int l, int r) {
  if (solve(l) == solve(r)) {
    for (int i = l; i <= r; i++) ans[i] = ans[l];
    return;
  }
  if (l + 1 >= r) return;
  int mid = (l + r) >> 1;
  Solve(l, mid);
  Solve(mid, r);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    ans[i] = -1;
    scanf("%d %d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  ans[n] = -1;
  dfs(1, 0);
  for (p = 1; p * p / (log(p) / log(2)) <= n; p++) solve(p);
  Solve(p, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
