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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) a = b;
}
const int N = 500 * 1000 + 228;
const int K = 1;
vector<int> g[N];
pair<int, int> dp[N];
int p[N];
vector<int> order;
void jfs(int v, int par = -1) {
  p[v] = par;
  for (int to : g[v]) {
    if (to != par) {
      jfs(to, v);
    }
  }
  order.push_back(v);
}
bool ca[N];
int re[N];
int doit(int k) {
  if (ca[k]) return re[k];
  ca[k] = 1;
  int down, mx1, mx2;
  for (int v : order) {
    dp[v] = {0, 1};
    down = 0;
    mx1 = 0, mx2 = 0;
    for (int to : g[v]) {
      down += dp[to].first;
      if (dp[to].second > mx1) {
        mx2 = mx1;
        mx1 = dp[to].second;
      } else if (dp[to].second > mx2)
        mx2 = dp[to].second;
    }
    if (mx1 + 1 == k) {
      dp[v] = {down + 1, 0};
    } else
      dp[v] = {down, mx1 + 1};
    if (mx1 && mx2 && mx1 + mx2 + 1 >= k && down + 1 > dp[v].first) {
      dp[v] = {down + 1, 0};
    }
  }
  re[k] = dp[1].first;
  return re[k];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  jfs(1);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < (int)g[i].size(); ++j) {
      if (g[i][j] == p[i]) {
        g[i].erase(g[i].begin() + j);
        break;
      }
    }
  }
  cout << n << '\n';
  for (int i = 2; i <= min(n, K); ++i) {
    cout << doit(i) << '\n';
  }
  int i = min(n, K) + 1;
  while (i <= n) {
    int l = i, r = n + 1;
    int x = doit(i);
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (doit(m) == x)
        l = m;
      else
        r = m;
    }
    for (int j = i; j <= l; ++j) {
      cout << x << '\n';
    }
    i = l + 1;
  }
  return 0;
}
