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("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
namespace IO {
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(long long &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { x = getchar(); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head), R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const long long &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.F), putchar(' '), _W(x.S);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head), putchar(sizeof...(tail) ? ' ' : '\n'), W(tail...);
}
}  // namespace IO
using namespace IO;
const int maxn = 1e5 + 50;
const int mod = 1e9 + 7;
long long qp(long long a, long long n) {
  long long res = 1;
  n %= mod - 1;
  if (n < 0) n += mod - 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
long long qp(long long a, long long n, int mod) {
  long long res = 1;
  n %= mod - 1;
  if (n < 0) n += mod - 1;
  while (n > 0) {
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
int n;
int res[maxn], mx[maxn];
int tot;
int head[maxn], nxt[maxn * 2], vt[maxn * 2], cnt;
void add(int u, int v) {
  vt[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
}
inline void dfs(int u, int pre, int k) {
  mx[u] = 0;
  int mxx = 0;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = vt[i];
    if (v == pre) continue;
    dfs(v, u, k);
    if (mx[v] > mx[u])
      mxx = mx[u], mx[u] = mx[v];
    else if (mx[v] > mxx)
      mxx = mx[v];
  }
  if (mx[u] + mxx + 1 >= k)
    mx[u] = 0, tot++;
  else
    mx[u]++;
}
inline void solve(int l, int r, int xl, int xr) {
  if (xl == xr) {
    for (int i = l; i <= r; ++i) res[i] = xl;
    return;
  }
  int mid = l + r >> 1;
  tot = 0;
  dfs(1, 0, mid);
  res[mid] = tot;
  if (l < mid) solve(l, mid - 1, tot, xr);
  if (r > mid) solve(mid + 1, r, xl, tot);
}
int main(int argc, char *argv[]) {
  R(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    R(u, v);
    add(u, v), add(v, u);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; ++i) W(res[i]);
  return 0;
}
