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 maxn = 1e5 + 10;
int dfq[maxn], Index;
int fa[maxn];
int head[maxn], nxt[maxn << 1], E[maxn << 1], Ecnt;
int f[maxn], g[maxn], maxl[maxn];
int n;
void Add_Edge(int x, int y) {
  nxt[++Ecnt] = head[x];
  head[x] = Ecnt;
  E[Ecnt] = y;
  nxt[++Ecnt] = head[y];
  head[y] = Ecnt;
  E[Ecnt] = x;
}
void DFS(int x, int father) {
  dfq[++Index] = x;
  fa[x] = father;
  for (int i = head[x]; i; i = nxt[i])
    if (E[i] != father) DFS(E[i], x);
}
void DP(int k) {
  if (f[k] != -1) return;
  memset(g, 0, sizeof(g));
  memset(maxl, 0, sizeof(maxl));
  for (int o = n; o >= 1; o--) {
    int x = dfq[o];
    int ml1 = 0, ml2 = 0;
    for (int i = head[x]; i; i = nxt[i])
      if (E[i] != fa[x]) {
        g[x] += g[E[i]];
        if (maxl[E[i]] > ml1)
          ml2 = ml1, ml1 = maxl[E[i]];
        else if (maxl[E[i]] > ml2)
          ml2 = maxl[E[i]];
      }
    if (ml1 + ml2 + 1 >= k)
      g[x]++;
    else
      maxl[x] = ml1 + 1;
  }
  f[k] = g[1];
}
void Solve(int l, int r) {
  if (f[l] == f[r]) {
    for (int i = l + 1; i < r; i++) f[i] = f[l];
    return;
  }
  int mid = (l + r) >> 1;
  DP(mid);
  DP(mid + 1);
  Solve(l, mid);
  Solve(mid + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; i++) scanf("%d%d", &x, &y), Add_Edge(x, y);
  DFS(1, 0);
  memset(f, 0xFF, sizeof(f));
  f[1] = n;
  Solve(1, n);
  for (int i = 1; i <= n; i++) printf("%d\n", f[i]);
  return 0;
}
