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 + 50, SQ = 320, INF = 0, MOD = 0;
int n, ans[N], G[N], dp[N], par[N], mx[N];
vector<int> adj[N], D;
void DFSB(int v, int p) {
  D.push_back(v);
  par[v] = p;
  for (auto u : adj[v]) {
    if (u == p) continue;
    DFSB(u, v);
  }
  return;
}
inline int get(int x) {
  if (G[x] >= 0) return G[x];
  for (int i = 1; i <= n; i++) dp[i] = 0, mx[i] = 0;
  for (auto v : D) {
    int mx1 = 0, mx2 = 0;
    for (auto u : adj[v]) {
      if (u == par[v]) continue;
      if (mx[u] > mx1) {
        mx2 = mx1;
        mx1 = mx[u];
      } else if (mx[u] > mx2)
        mx2 = mx[u];
      dp[v] += dp[u];
    }
    if (mx1 + mx2 + 1 >= x) dp[v]++, mx1 = -1;
    mx[v] = mx1 + 1;
  }
  G[x] = dp[1];
  return dp[1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) G[i] = -1;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  DFSB(1, 0);
  reverse(D.begin(), D.end());
  for (int i = 1; i <= min(n, SQ); i++) ans[i] = get(i);
  for (int i = SQ + 1; i <= n; i++) {
    int last = get(i);
    int l = i - 1, r = n + 1;
    while (r - l > 1) {
      int mid = (r + l) / 2;
      if (get(mid) == last)
        l = mid;
      else
        r = mid;
    }
    for (int j = i; j <= l; j++) ans[j] = last;
    i = l;
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << '\n';
  return 0;
}
