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 sqn = 1100;
int n, m = 0, dp[100069], fh[100069], sq[100069];
pair<int, int> ed[200069];
vector<int> al[100069];
bitset<100069> vtd;
void rtd(int x) {
  int i, sz = al[x].size(), l;
  vector<int> v;
  vtd[x] = 1;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      rtd(l);
      v.push_back(l);
    }
  }
  sz = v.size();
  for (i = 0; i < sz; i++) {
    l = v[i];
    m++;
    ed[m] = {x, l};
  }
  m++;
  ed[m] = {x, 0};
}
inline int slv(int x) {
  int i, k, l, mx[2] = {0, 0};
  for (i = 1; i <= n; i++) {
    dp[i] = 0;
    fh[i] = 0;
  }
  for (i = 1; i <= m; i++) {
    k = ed[i].first;
    l = ed[i].second;
    if (l) {
      dp[k] += dp[l];
      fh[k] = max(fh[k], fh[l] + 1);
      mx[1] = max(mx[1], min(mx[0], fh[l] + 1));
      mx[0] = max(mx[0], fh[l] + 1);
    } else {
      if (mx[0] + mx[1] + 1 >= x) {
        dp[k]++;
        fh[k] = -1;
      }
      mx[0] = 0;
      mx[1] = 0;
    }
  }
  return dp[1];
}
int main() {
  int i, k, l, lh, rh, md, zz;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &k, &l);
    al[k].push_back(l);
    al[l].push_back(k);
  }
  rtd(1);
  for (i = 1; i <= min(n / sqn, n); i++) {
    for (lh = 1, rh = n / i; lh <= rh;) {
      md = (lh + rh) / 2;
      if (slv(md) >= i) {
        zz = md;
        lh = md + 1;
      } else {
        rh = md - 1;
      }
    }
    sq[zz] = i;
  }
  for (i = n - 1; i; i--) {
    sq[i] = max(sq[i], sq[i + 1]);
  }
  for (i = 1; i <= min(sqn, n); i++) {
    sq[i] = slv(i);
  }
  for (i = 1; i <= n; i++) {
    printf("%d\n", sq[i]);
  }
}
