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>
inline int gi() {
  char c = getchar();
  int x = 0;
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  return x;
}
const int N = 100005;
int head[N], nxt[N << 1], to[N << 1], n, k, cnt, f[N], fa[N], st[N], tp;
void addedge(int u, int v, int now) {
  nxt[now] = head[u], head[u] = now, to[now] = v;
}
void dfs(int u) {
  for (int e = head[u]; e; e = nxt[e])
    if (to[e] != fa[u]) fa[to[e]] = u, dfs(to[e]);
  st[++tp] = u;
}
int solve(int k) {
  int cnt = 0;
  for (int i = 1; i <= n; ++i) f[i] = 1;
  for (int i = 1; i < n; ++i) {
    int u = st[i];
    if (f[u] == -1 || f[fa[u]] == -1) continue;
    if (f[fa[u]] + f[u] >= k)
      ++cnt, f[fa[u]] = -1;
    else if (f[u] + 1 > f[fa[u]])
      f[fa[u]] = f[u] + 1;
  }
  return cnt;
}
int main() {
  n = gi();
  printf("%d\n", n);
  for (int i = 1; i < n; ++i) {
    int u = gi(), v = gi();
    addedge(u, v, i * 2 - 1);
    addedge(v, u, i * 2);
  }
  dfs(1);
  int m = sqrt(1.0 * n * log(n) / log(2));
  for (int i = 2; i <= m; ++i) printf("%d\n", solve(i));
  for (int i = m + 1; i <= n;) {
    int now = solve(i);
    int l = i + 1, r = n;
    while (l <= r) {
      int mid = l + r >> 1;
      solve(mid) == now ? l = mid + 1 : r = mid - 1;
    }
    for (; i < l; ++i) printf("%d\n", now);
  }
}
