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 = 100005;
int n, s;
int a, b;
int K, ret;
bool vis[N];
int maxPath[N];
int ans[N];
vector<int> V[N];
int preorder[N];
int preTime;
void dfs(int w) {
  preorder[++preTime] = w;
  vis[w] = true;
  for (int i = V[w].size() - 1; i >= 0; i--) {
    int u = V[w][i];
    if (vis[u]) {
      swap(V[w][i], V[w].back());
      V[w].pop_back();
      continue;
    }
    dfs(u);
  }
}
int solve(int k) {
  ret = 0;
  K = k;
  for (int i = n; i >= 1; i--) {
    int w = preorder[i];
    maxPath[w] = 1;
    bool done = false;
    for (int u : V[w]) {
      if (maxPath[w] + maxPath[u] >= K) {
        maxPath[w] = 0;
        ret++;
        done = true;
        break;
      }
      maxPath[w] = max(maxPath[w], maxPath[u] + 1);
    }
    if (!done && maxPath[w] >= K) {
      maxPath[w] = 0;
      ret++;
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  int l = 1;
  int w = n;
  while (w) {
    w >>= 1;
    l++;
  }
  s = (int)sqrt(n * l);
  for (int i = 2; i <= n; i++) {
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  dfs(1);
  ans[1] = n;
  for (int i = 2; i <= s; i++) {
    ans[i] = solve(i);
  }
  int max_ans = n / (s + 1);
  int last = s;
  vector<int> nexts;
  nexts.push_back(s);
  set<pair<int, int> > visited;
  for (int i = max_ans; i >= 0; i--) {
    auto it = visited.upper_bound({-i, n});
    auto it2 = it;
    int low = last + 1;
    if (it != visited.begin()) {
      --it2;
      low = max(low, it2->second + 1);
    }
    int high = n;
    if (it != visited.end()) {
      high = it->second - 1;
    }
    int res = low - 1;
    while (low <= high) {
      int mid = (low + high) / 2;
      int paths = solve(mid);
      visited.insert({-paths, mid});
      if (paths >= i) {
        res = mid;
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
    nexts.push_back(max(nexts.back(), res));
    last = res;
  }
  int cnt = 0;
  for (int i = nexts.size() - 1; i >= 1; i--) {
    for (int j = nexts[i]; j > nexts[i - 1]; j--) {
      ans[j] = cnt;
    }
    cnt++;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
