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;
mt19937 rnd(time(nullptr));
const long long INF = 1e18, N = 1e5 + 10, B = 317;
vector<int> g[N], g2[N];
int ans_vec[N], depth[N], m1[N], m2[N], ans = 0, used[N], n, tout[N], t = 0;
void dfs(int v, int pr) {
  for (int u : g[v]) {
    if (u != pr) {
      dfs(u, v);
    }
  }
  tout[v] = t;
  t++;
}
int get_vl(int k) {
  ans = 0;
  for (int v = 0; v < n; v++) {
    m1[v] = 0;
    m2[v] = 0;
    for (int u : g2[v]) {
      if (depth[u] > m1[v]) {
        m2[v] = m1[v];
        m1[v] = depth[u];
      } else if (depth[u] > m2[v]) {
        m2[v] = depth[u];
      }
    }
    if (m1[v] + m2[v] + 1 >= k) {
      ans++;
      depth[v] = 0;
    } else {
      depth[v] = m1[v] + 1;
    }
  }
  return ans;
}
int my_min(int a, int b) {
  if (a > b) {
    return b;
  }
  return a;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    for (int j : g[i]) {
      int i1 = tout[i], j1 = tout[j];
      if (i1 < j1) {
        continue;
      }
      g2[i1].push_back(j1);
    }
  }
  for (int k = 1; k <= my_min(n, B); k++) {
    ans_vec[k] = get_vl(k);
  }
  for (int i = B + 1; i <= n;) {
    int l = i, r = n + 1;
    int bg = get_vl(i);
    while (r - l > 1) {
      int mid = (l + r) / 2;
      int y = get_vl(mid);
      if (y == bg) {
        l = mid;
      } else {
        r = mid;
      }
    }
    for (int j = i; j <= l; j++) {
      ans_vec[j] = bg;
    }
    i = l + 1;
  }
  for (int i = 1; i <= n; i++) {
    cout << ans_vec[i] << '\n';
  }
  cout << '\n';
  return 0;
}
