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;
int Begin[N], Next[N], to[N], e;
void add(int u, int v) { to[++e] = v, Next[e] = Begin[u], Begin[u] = e; }
vector<int> G[N];
int dep[N], mxl[N], clk;
void init(int o, int fa) {
  int u = ++clk;
  mxl[u] = 1;
  for (int ver : G[o]) {
    if (ver == fa) continue;
    int v = clk + 1;
    add(u, clk + 1);
    init(ver, o);
    mxl[u] = max(mxl[u], mxl[v]);
    mxl[u] = max(mxl[u], dep[u] + dep[v] + 1);
    dep[u] = max(dep[u], dep[v]);
  }
  dep[u]++;
}
int k;
int len[N], cur;
void DFS(int o) {
  if (mxl[o] < k) {
    len[o] = dep[o];
    return;
  }
  int mx = 0;
  for (int i = Begin[o]; i; i = Next[i]) {
    int v = to[i];
    DFS(v);
    if (mx != -1) {
      if (mx + len[v] + 1 >= k)
        ++cur, mx = -1;
      else if (len[v] > mx)
        mx = len[v];
    }
  }
  len[o] = mx + 1;
}
int calc(int x) {
  k = x;
  cur = 0;
  DFS(1);
  return cur;
}
int n;
int ans[N];
void work(int L, int R, int lval, int rval) {
  if (R - L <= 1) {
    ans[L] = lval, ans[R] = rval;
    return;
  }
  if (lval == rval) {
    for (int i = L; i <= R; i++) ans[i] = lval;
    return;
  }
  int M = (L + R + 1) / 2;
  int x = calc(M);
  work(L, M, lval, x), work(M, R, x, rval);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  init(1, 0);
  work(1, n, n, calc(n));
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
