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;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 111111;
vector<int> a[N];
int par[N];
vector<int> order;
void dfs_par(int v) {
  for (int x : a[v]) {
    if (x == par[v]) {
      continue;
    }
    par[x] = v;
    dfs_par(x);
  }
  order.push_back(v);
}
struct TwoMax {
  int x, y;
  TwoMax() { x = y = 0; }
  void relax(int z) {
    if (z > x) {
      y = x;
      x = z;
    } else if (z > y) {
      y = z;
    }
  }
};
void maxPath(int n, int len, int& ans) {
  vector<int> tmp(n);
  for (auto v : order) {
    TwoMax tm;
    for (int x : a[v]) {
      if (x == par[v]) {
        continue;
      }
      tm.relax(tmp[x]);
    }
    if (tm.x + tm.y >= len) {
      ++ans;
    } else {
      tmp[v] = tm.x + 1;
    }
  }
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int n = nxt();
  for (int i = 0; i < n - 1; ++i) {
    int u = nxt() - 1, v = nxt() - 1;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  par[0] = -1;
  dfs_par(0);
  vector<int> ans(n);
  function<void(int, int, int, int)> fill = [&](int l, int r, int u, int d) {
    if (u == d) {
      for (int i = l; i < r; ++i) {
        ans[i] = d;
      }
      return;
    }
    if (l >= r) {
      return;
    }
    int m = (l + r) / 2;
    maxPath(n, m, ans[m]);
    fill(l, m, u, ans[m]);
    fill(m + 1, r, ans[m], d);
  };
  fill(0, n, n, 0);
  for (int i = 0; i < n; ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
