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;
struct E {
  int to, nex;
} e[N * 2];
int head[N], ecnt;
void adde(int fr, int to) {
  e[++ecnt] = (E){to, head[fr]};
  head[fr] = ecnt;
}
void addedge(int fr, int to) {
  adde(fr, to);
  adde(to, fr);
}
int n, f[N], res;
void dfs(int o, int fa, int mid) {
  int mx1 = 0, mx2 = 0;
  for (int j = head[o]; j; j = e[j].nex)
    if (e[j].to != fa) {
      dfs(e[j].to, o, mid);
      if (mx1 < f[e[j].to])
        mx2 = mx1, mx1 = f[e[j].to];
      else
        mx2 = max(mx2, f[e[j].to]);
    }
  if (mx1 + mx2 + 1 >= mid)
    res++, f[o] = 0;
  else
    f[o] = mx1 + 1;
}
int ans[N];
void solve(int l, int r, int ql, int qr) {
  if (l > r || ql > qr) return;
  if (l == r) {
    for (int i = ql; i <= qr; i++) ans[i] = l;
    return;
  }
  int mid = ql + qr >> 1;
  res = 0;
  dfs(1, 0, mid);
  ans[mid] = res;
  solve(l, res, mid + 1, qr);
  solve(res, r, ql, mid - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
  }
  solve(0, n, 1, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
