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 maxn = 1000 * 100 + 5;
int mx[maxn], mxx[maxn], pr[maxn], h[maxn], n;
vector<int> aj[maxn];
pair<int, int> v[maxn];
bool mrk[maxn];
int ans(int k) {
  int ANS = 0;
  for (int i = 0; i < n; i++) {
    mx[i] = 1;
    mxx[i] = 1;
  }
  for (int i = n - 1; i >= 0; i--) {
    int u = v[i].second;
    if (mxx[u] + mx[u] > k) {
      ANS++;
      mx[u] = 0;
      mxx[u] = 0;
    }
    if (u != 0) {
      int v = pr[u];
      (mxx[v]) = max((mxx[v]), (mx[u] + 1));
      if (mx[v] < mxx[v]) swap(mxx[v], mx[v]);
    }
  }
  return ANS;
}
void pre(int v) {
  mrk[v] = true;
  for (int i = 0; i < (int)(aj[v]).size(); i++) {
    int u = aj[v][i];
    if (!mrk[u]) {
      pr[u] = v;
      h[u] = h[v] + 1;
      pre(u);
    }
  }
  return;
}
int32_t main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    aj[v].push_back(u);
    aj[u].push_back(v);
  }
  pre(0);
  for (int i = 0; i < n; i++) v[i] = {h[i], i};
  sort(v, v + n);
  for (int i = 1; i <= n; i++) {
    int x = ans(i);
    if (i <= 1300)
      printf("%d \n", x);
    else {
      int l = i, r = n + 1;
      while (r - l > 1) {
        int mid = (r + l) / 2;
        if (ans(mid) == x)
          l = mid;
        else
          r = mid;
      }
      for (int j = i; j <= l; j++) printf("%d \n", x);
      i = l;
    }
  }
  return false;
}
