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 mxn = 100010, siz = 1500;
int n, m, hd[mxn], head[mxn], f[mxn], g[mxn];
struct ed {
  int to, nxt;
} edge[mxn], edg[mxn << 1];
void addedge(int u, int v) { edge[++m] = (ed){v, head[u]}, head[u] = m; }
void addedg(int u, int v) {
  edg[++m] = (ed){v, hd[u]}, hd[u] = m;
  edg[++m] = (ed){u, hd[v]}, hd[v] = m;
}
void dfss(int u, int fa) {
  for (int i = hd[u], v; i; i = edg[i].nxt)
    if ((v = edg[i].to) != fa) dfss(v, u), addedge(u, v);
}
void dfs(int u, int k) {
  f[u] = 0;
  int fir = 0, sec = 0;
  for (int i = head[u], v; i; i = edge[i].nxt) {
    dfs(v = edge[i].to, k), f[u] += f[v];
    if (g[v] > fir)
      sec = fir, fir = g[v];
    else
      sec = max(sec, g[v]);
  }
  if (fir + sec + 1 >= k)
    ++f[u], g[u] = 0;
  else
    g[u] = fir + 1;
}
int solve(int k) {
  dfs(1, k);
  return f[1];
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i) scanf("%d%d", &x, &y), addedg(x, y);
  m = 0;
  dfss(1, 0);
  if (n <= siz) {
    for (int i = 1; i <= n; ++i) printf("%d\n", solve(i));
    return 0;
  }
  for (int i = 1; i <= siz; ++i) printf("%d\n", solve(i));
  for (int i = n / (siz + 1), cur = siz + 1, l, r; i >= 0; --i) {
    for (l = cur, r = n; l <= r;)
      if (solve(((l + r) >> 1)) < i)
        r = ((l + r) >> 1) - 1;
      else
        l = ((l + r) >> 1) + 1;
    for (int j = cur; j <= r; ++j) printf("%d\n", i);
    cur = l;
  }
  return 0;
}
