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;
const int maxn = 5e5 + 7, mod = 1e9 + 7, MAXN = 1e6 + 7;
const double eps = 1e-9;
const long long inf = 1e18;
mt19937 rnd(time(0));
struct req {
  int l, r, x;
};
int n;
vector<int> gr[maxn];
pair<int, int> down[maxn];
int up[maxn];
int prev228[maxn];
void relax(pair<int, int> &a, pair<int, int> b) {
  if (a.first < b.first) {
    a.second = a.first;
    a.first = b.first;
  } else if (a.second < b.first) {
    a.second = b.first;
  }
}
pair<int, int> dfs(int u, int pr) {
  prev228[u] = pr;
  pair<int, int> t = {-1, -1};
  for (auto v : gr[u]) {
    if (v == pr) continue;
    relax(t, dfs(v, u));
  }
  t.first++;
  t.second++;
  return down[u] = t;
}
void dfs1(int u, int pr) {
  if (u != pr) {
    up[u] = up[pr] + 1;
    if (down[pr].first - 1 == down[u].first) {
      up[u] = max(up[u], down[pr].second + 1);
    } else {
      up[u] = max(up[u], down[pr].first + 1);
    }
  }
  for (auto v : gr[u]) {
    if (v == pr) continue;
    dfs1(v, u);
  }
}
vector<int> depth[maxn];
int ptr[maxn];
bool check(int a, int b, int l) {
  int ok = 0;
  for (int i = 0; i < min((int)depth[b].size(), 3); ++i) {
    ok += (depth[b][i] >= l);
  }
  ok += (up[a] >= l);
  int ok1 = ok;
  bool used = false;
  for (int i = 0; i < min((int)depth[a].size(), 4); ++i) {
    if (!used && depth[a][i] == down[b].first + 1) {
      used = true;
      continue;
    }
    ok += (depth[a][i] >= l);
  }
  return (ok1 && (ok - ok1) && ok >= 3);
}
int get(int a, int b, int l) {
  while (ptr[a] >= 0 && depth[a][ptr[a]] < l) --ptr[a];
  while (ptr[b] >= 0 && depth[b][ptr[b]] < l) --ptr[b];
  int stick = ptr[a] + ptr[b] + 2;
  if (up[a] >= l) ++stick;
  if (down[b].first + 1 >= l) --stick;
  return stick;
}
int ans[maxn];
vector<req> odd, even;
vector<pair<int, int> > value[maxn];
void get(vector<req> &q, int t) {
  for (auto &v : q) {
    v.l = min(v.l, n);
    v.r = min(v.r, n);
    value[v.l].emplace_back(v.x, 1);
    value[v.r + 1].emplace_back(v.x, -1);
  }
  multiset<int> mx;
  for (int i = 0; i <= n; ++i) {
    for (auto v : value[i]) {
      if (v.second == -1)
        mx.erase(mx.find(v.first));
      else
        mx.insert(v.first);
    }
    if (mx.size() && i % 2 == t) ans[i] = max(ans[i], *(--mx.end()));
  }
  for (auto v : q) {
    value[v.l].clear();
    value[v.r + 1].clear();
  }
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b, --a, --b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  dfs(0, 0);
  dfs1(0, 0);
  for (int i = 0; i < n; ++i) {
    for (auto v : gr[i]) {
      if (v == prev228[i]) continue;
      depth[i].push_back(down[v].first + 1);
    }
    ptr[i] = (int)depth[i].size() - 1;
    sort(depth[i].begin(), depth[i].end());
    reverse(depth[i].begin(), depth[i].end());
  }
  vector<pair<int, int> > E;
  for (int i = 1; i < n; ++i) {
    E.emplace_back(prev228[i], i);
  }
  for (int i = 2; i <= n; i += 2) {
    vector<pair<int, int> > nw;
    for (auto v : E) {
      if (check(v.first, v.second, i / 2)) {
        nw.push_back(v);
      }
    }
    swap(nw, E);
    for (auto v : E) {
      ans[i] = max(ans[i], get(v.first, v.second, i / 2));
    }
  }
  odd.push_back({0, n, 1});
  even.push_back({0, n, 1});
  for (int u = 0; u < n; ++u) {
    vector<int> value;
    for (auto v : depth[u]) value.push_back(v);
    value.push_back(up[u]);
    value.push_back(0);
    sort(value.begin(), value.end());
    map<int, int> cnt;
    for (auto v : value) cnt[v]++;
    value.erase(unique(value.begin(), value.end()), value.end());
    vector<int> suf((int)value.size() + 1);
    for (int i = (int)value.size() - 1; i >= 0; --i) {
      suf[i] = suf[i + 1] + cnt[value[i]];
    }
    for (int i = 1; i < (int)value.size(); ++i) {
      odd.push_back({value[i - 1] * 2, value[i] * 2, suf[i]});
      int l = value[i - 1] + 1, r = value[i] - 1;
      l *= 2, r *= 2;
      ++l, ++r;
      if (l <= r) even.push_back({l, r, suf[i]});
      ans[min(n, value[i] * 2 + 1)] =
          max(ans[min(n, value[i] * 2 + 1)], 1 + suf[i + 1]);
    }
  }
  get(odd, 0);
  get(even, 1);
  for (int i = 0; i < n; ++i) {
    ans[1] = max(ans[1], (int)gr[i].size() + 1);
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int t = 1;
  for (int i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}