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 INF = 1e9 + 7;
const long long LL_INF = 1ll * INF * INF;
const int MAX_N = 300000 + 7;
const int mod = INF;
template <typename T>
inline void addmod(T& a, const long long& b, const int& MOD = mod) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
int f[MAX_N];
vector<int> adj[MAX_N];
int n;
int pa[MAX_N];
vector<int> rev;
void dfs(int u, int r = -1) {
  pa[u] = r;
  for (int v : adj[u])
    if (v != r) {
      dfs(v, u);
    }
  rev.push_back(u);
}
int pre_calc[MAX_N];
int calc(int k) {
  if (pre_calc[k] >= 0) return pre_calc[k];
  int cnt = 0;
  for (int u : rev) {
    static int child_lens[2], c_sz;
    c_sz = 0;
    for (int v : adj[u])
      if (v != pa[u]) {
        if (c_sz < 2) {
          child_lens[c_sz++] = f[v];
          if (c_sz == 2 && child_lens[0] < child_lens[1])
            swap(child_lens[0], child_lens[1]);
        } else {
          if (f[v] > child_lens[0]) {
            child_lens[1] = child_lens[0];
            child_lens[0] = f[v];
          } else if (f[v] > child_lens[1]) {
            child_lens[1] = f[v];
          }
        }
      }
    int mx_len = 1;
    if (c_sz >= 1) mx_len = max(mx_len, child_lens[0] + 1);
    if (c_sz >= 2) mx_len = max(mx_len, child_lens[0] + 1 + child_lens[1]);
    if (mx_len >= k) {
      ++cnt;
      f[u] = 0;
    } else {
      f[u] = 1;
      if (c_sz > 0) f[u] += child_lens[0];
    }
  }
  return pre_calc[k] = cnt;
}
void solve() {
  cin >> n;
  for (int i = (1), _b = (n); i < _b; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  fill_n(pre_calc, n + 1, -1);
  const int M = min(n, (int)sqrt(n * log2(n)));
  for (int i = 1; i <= M; ++i) {
    cout << calc(i) << '\n';
  }
  for (int i = M + 1; i <= n;) {
    int lo = i + 1, hi = n, last = i;
    int cur = calc(i);
    if (!cur) {
      while (i <= n) {
        cout << cur << '\n';
        ++i;
      }
      break;
    }
    while (lo <= hi) {
      int mid = (lo + hi) >> 1;
      if (calc(mid) == cur) {
        last = mid;
        lo = mid + 1;
      } else
        hi = mid - 1;
    }
    while (i <= last) {
      cout << cur << '\n';
      ++i;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}
