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;
int cnt, n, a, b, head[120000], dp[120000], ans[120000];
struct wxp {
  int nxt, to;
} e[230000];
void add(int u, int v) {
  e[++cnt] = (wxp){head[u], v}, head[u] = cnt;
  e[++cnt] = (wxp){head[v], u}, head[v] = cnt;
}
void dfs(int u, int fa, int snd) {
  dp[u] = 0;
  int mx1 = 0, mx2 = 0;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(v, u, snd);
    if (dp[v] > mx1)
      mx2 = mx1, mx1 = dp[v];
    else
      mx2 = max(mx2, dp[v]);
  }
  if (mx1 + mx2 + 1 >= snd)
    cnt++, dp[u] = 0;
  else
    dp[u] = mx1 + 1;
}
void sol(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = (l + r) / 2;
  cnt = 0, dfs(1, -1, mid), ans[mid] = cnt;
  sol(l, mid - 1, cnt, R), sol(mid + 1, r, L, cnt);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d", &a, &b), add(a, b);
  sol(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
}
