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;
template <class T>
inline T re() {
  T N = 0;
  char c = getchar();
  bool neg = 0;
  for (; c < '0' || c > '9'; c = getchar()) neg |= c == '-';
  for (; c >= '0' && c <= '9'; c = getchar()) N = (N << 3) + (N << 1) + c - '0';
  return neg ? -N : N;
}
const int MX = 1e5;
int n;
vector<int> adj[MX + 5];
int par[MX + 5], at[MX + 5], ord[MX + 5];
int np[MX + 5], mxl[MX + 5];
int fst[MX + 5], snd[MX + 5];
void post(int u, int prv = 0) {
  static int tme = 0;
  par[u] = prv;
  for (int nx : adj[u])
    if (nx != prv) post(nx, u);
  at[++tme] = u;
  ord[u] = tme;
}
int solve(int k) {
  fill(np + 1, np + n + 1, 0);
  fill(mxl + 1, mxl + n + 1, 0);
  fill(fst + 1, fst + n + 1, 0);
  fill(snd + 1, snd + n + 1, 0);
  for (int i = 1, u; i <= n; i++) {
    u = at[i];
    mxl[u] = fst[u] + 1;
    if (mxl[u] + snd[u] >= k) {
      mxl[u] += snd[u];
    }
    if (mxl[u] >= k) {
      mxl[u] = 0;
      np[u]++;
    }
    if (par[u]) {
      np[par[u]] += np[u];
      if (mxl[u] > fst[par[u]]) {
        snd[par[u]] = fst[par[u]], fst[par[u]] = mxl[u];
      } else if (mxl[u] > snd[par[u]]) {
        snd[par[u]] = mxl[u];
      }
    }
  }
  return np[1];
}
int ans[MX + 5];
void get(int l, int r, int optl, int optr) {
  if (l > r) return;
  if (optl == optr) {
    for (int i = l; i <= r; i++) ans[i] = optl;
  } else {
    int md = (l + r) >> 1;
    ans[md] = solve(md);
    get(l, md - 1, ans[md], optr);
    get(md + 1, r, optl, ans[md]);
  }
}
int main() {
  n = re<int>();
  for (int i = 1, u, v; i < n; i++) {
    u = re<int>();
    v = re<int>();
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  post(1);
  get(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
