Problem: Let G be a simple graph. Let W be a non-empty subset of vertices. Then W is almost-k-uniform if for each pair of distinct vertices u,v ∈ W the distance between u and v is either k or k+1.

You are given a tree on n vertices. For each i between 1 and n, find the maximum size of an almost-i-uniform set.

Input

The first line contains a single integer n (2 ≤ n ≤ 5 ⋅ 10^5) – the number of vertices of the tree.

Then n-1 lines follows, the i-th of which consisting of two space separated integers u_i, v_i (1 ≤ u_i, v_i ≤ n) meaning that there is an edge between vertices u_i and v_i. 

It is guaranteed that the given graph is tree. 

Output

Output a single line containing n space separated integers a_i, where a_i is the maximum size of an almost-i-uniform set.

Examples

Input


5
1 2
1 3
1 4
4 5


Output


4 3 2 1 1


Input


6
1 2
1 3
1 4
4 5
4 6


Output


4 4 2 1 1 1

Note

Consider the first example. 

  * The only maximum almost-1-uniform set is \{1, 2, 3, 4\}. 
  * One of the maximum almost-2-uniform sets is or \{2, 3, 5\}, another one is \{2, 3, 4\}. 
  * A maximum almost-3-uniform set is any pair of vertices on distance 3. 
  * Any single vertex is an almost-k-uniform set for k ≥ 1. 



In the second sample there is an almost-2-uniform set of size 4, and that is \{2, 3, 5, 6\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template <class t>
using vvc = vc<vc<t>>;
using pi = pair<ll, ll>;
using vi = vc<ll>;
template <class t, class u>
ostream& operator<<(ostream& os, const pair<t, u>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class t>
ostream& operator<<(ostream& os, const vc<t>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
using uint = unsigned;
using ull = unsigned long long;
template <class t, size_t n>
ostream& operator<<(ostream& os, const array<t, n>& first) {
  return os << vc<t>(first.begin(), first.end());
}
template <ll i, class T>
void print_tuple(ostream&, const T&) {}
template <ll i, class T, class H, class... Args>
void print_tuple(ostream& os, const T& t) {
  if (i) os << ",";
  os << get<i>(t);
  print_tuple<i + 1, T, Args...>(os, t);
}
template <class... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
  os << "{";
  print_tuple<0, tuple<Args...>, Args...>(os, t);
  return os << "}";
}
void print(ll x, ll suc = 1) {
  cout << x;
  if (suc == 1) cout << "\n";
  if (suc == 2) cout << " ";
}
ll read() {
  ll i;
  cin >> i;
  return i;
}
vi readvi(ll n, ll off = 0) {
  vi v(n);
  for (ll i = ll(0); i < ll(n); i++) v[i] = read() + off;
  return v;
}
template <class T>
void print(const vector<T>& v, ll suc = 1) {
  for (ll i = ll(0); i < ll(v.size()); i++)
    print(v[i], i == ll(v.size()) - 1 ? suc : 2);
}
string readString() {
  string s;
  cin >> s;
  return s;
}
template <class T>
T sq(const T& t) {
  return t * t;
}
void yes(bool ex = true) {
  cout << "Yes" << endl;
  if (ex) exit(0);
}
void no(bool ex = true) {
  cout << "No" << endl;
  if (ex) exit(0);
}
void possible(bool ex = true) {
  cout << "Possible" << endl;
  if (ex) exit(0);
}
void impossible(bool ex = true) {
  cout << "Impossible" << endl;
  if (ex) exit(0);
}
constexpr ll ten(ll n) { return n == 0 ? 1 : ten(n - 1) * 10; }
const ll infLL = LLONG_MAX / 3;
const ll inf = infLL;
ll topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
ll topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
ll botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
ll botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
ll popcount(signed t) { return __builtin_popcount(t); }
ll popcount(ll t) { return __builtin_popcountll(t); }
bool ispow2(ll i) { return i && (i & -i) == i; }
ll mask(ll i) { return (ll(1) << i) - 1; }
bool inc(ll first, ll second, ll c) { return first <= second && second <= c; }
template <class t>
void mkuni(vc<t>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
ll rand_int(ll l, ll r) {
  static random_device rd;
  static mt19937_64 gen(rd());
  return uniform_int_distribution<ll>(l, r)(gen);
}
template <class t>
ll lwb(const vc<t>& v, const t& first) {
  return lower_bound(v.begin(), v.end(), first) - v.begin();
}
const ll nmax = 500010;
vi t[nmax];
ll len[nmax], dia;
using E = tuple<ll, ll, ll>;
vc<E> es;
ll par[nmax];
void dfs1(ll v, ll p) {
  par[v] = p;
  for (auto c : t[v])
    if (c != p) {
      dfs1(c, v);
      chmax(len[v], len[c]);
    }
  len[v]++;
  if (p != -1) es.emplace_back(len[v], v, p);
}
void dfs2(ll v, ll p, ll d) {
  if (p != -1) es.emplace_back(d, p, v);
  vi z{d};
  for (auto c : t[v])
    if (c != p) {
      z.push_back(len[c]);
    }
  sort(z.begin(), z.end(), greater<ll>());
  chmax(dia, z[0]);
  for (auto c : t[v])
    if (c != p) {
      ll w = z[0];
      if (w == len[c]) w = z[1];
      dfs2(c, v, w + 1);
    }
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  ll n;
  cin >> n;
  for (ll _ = ll(0); _ < ll(n - 1); _++) {
    ll first, second;
    cin >> first >> second;
    first--;
    second--;
    t[first].push_back(second);
    t[second].push_back(first);
  }
  dfs1(0, -1);
  dfs2(0, -1, 0);
  sort(es.begin(), es.end(), greater<E>());
  vi e(n + 2, -1), oe(n + 2, -1), eo(n + 2, -1);
  vi cnt(n), pre(n, -1), nei(n, 0);
  for (auto www : es) {
    void(0);
    ll k, u, v;
    tie(k, u, v) = www;
    {
      if (cnt[v] >= 1) {
        chmax(e[k * 2], cnt[v] + 1);
      }
    }
    {
      if (cnt[v] >= 2 && pre[v] > k) {
        chmax(oe[k * 2 + 2], cnt[v] + 1);
      }
    }
    {
      ll z = nei[v];
      if (par[v] != -1) chmax(z, cnt[par[v]]);
      if (z >= 2) {
        z--;
        chmax(eo[k * 2 + 1], z + cnt[v]);
      }
    }
    cnt[v]++;
    pre[v] = k;
    if (par[v] != -1) chmax(nei[par[v]], cnt[v]);
  }
  void(0);
  void(0);
  void(0);
  for (ll i = ll(n + 1) - 1; i >= ll(0); i--) chmax(e[i], e[i + 1]);
  for (ll i = ll(n + 1) - 1; i >= ll(0); i--) chmax(oe[i], oe[i + 1]);
  for (ll i = ll(n + 1) - 1; i >= ll(0); i--) chmax(eo[i], eo[i + 1]);
  vi ans;
  for (ll i = ll(2); i < ll(n + 2); i++) {
    ll w = 1;
    if (i - 1 <= dia) w = 2;
    if (i % 2 == 0) {
      chmax(w, e[i]);
      chmax(w, oe[i]);
    } else {
      chmax(w, eo[i]);
      chmax(w, e[i - 1]);
    }
    ans.push_back(w);
  }
  for (ll i = ll(0); i < ll(n); i++) chmax(ans[0], (ll)t[i].size() + 1);
  print(ans);
}