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 = 1e6 + 10;
const int mod = 1e9 + 7;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
vector<int> g[N];
int mx1[N], mx2[N], dfn[N], tot, z[N];
void dfs(int x, int fa) {
  for (int to : g[x]) {
    if (to == fa) continue;
    dfs(to, x);
  }
  z[x] = fa;
  dfn[++tot] = x;
}
int solve(int n, int k) {
  int res = 0;
  for (int i = 1; i <= n; i++) mx1[i] = mx2[i] = 0;
  for (int i = 1; i <= n; i++) {
    int x = dfn[i], y = z[x];
    if (mx1[x] + mx2[x] + 1 >= k) {
      res++;
    } else {
      int tmp = mx1[x] + 1;
      if (tmp >= mx1[y])
        mx2[y] = mx1[y], mx1[y] = tmp;
      else if (tmp > mx2[y])
        mx2[y] = tmp;
    }
  }
  return res;
}
int main() {
  int n = read();
  int block = min(n, 200);
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  vector<int> ans(n + 1);
  for (int i = 1; i <= block; i++) {
    ans[i] = solve(n, i);
  }
  for (int i = block + 1, j; i <= n; i = j + 1) {
    int o = solve(n, i);
    int l = i, r = n;
    while (l <= r) {
      int mid = (l + r) >> 1;
      int z = solve(n, mid);
      if (z == o) {
        j = mid;
        l = mid + 1;
      } else
        r = mid - 1;
    }
    for (int t = i; t <= j; t++) ans[t] = o;
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
}
