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>
const int N = 100010;
const int buff_size = 1e6;
char gc() {
  static char buf[buff_size], *p1 = buf, *p2 = buf;
  return p1 == p2 &&
                 (p2 = (p1 = buf) + fread(buf, 1, buff_size, stdin), p1 == p2)
             ? EOF
             : *p1++;
};
inline void pc(const char &c = 0) {
  static char buf[buff_size], *p = buf;
  if (p - buf == buff_size || !c) fwrite(buf, 1, p - buf, stdout), p = buf;
  *p++ = c;
}
template <class Tp = int>
inline void write(Tp x) {
  if (x < 0) x = -x, pc('-');
  static int sta[35];
  int top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) pc(sta[--top] ^ 48);
}
template <class Tp = int>
inline Tp read() {
  Tp w = Tp();
  bool f = 0;
  char ch = gc();
  while (!isdigit(ch)) f |= ch == '-', ch = gc();
  while (isdigit(ch)) w = (w << 3) + (w << 1) + (ch ^ 48), ch = gc();
  return f ? -w : w;
}
using namespace std;
int n, S, s[N], ans[N], res;
struct llz {
  int link, next;
} e[N << 1];
int tot, head[N];
inline void add(int u, int v) {
  e[++tot] = (llz){v, head[u]};
  head[u] = tot;
}
void dfs(int p, int f, int NEKO) {
  s[p] = 0;
  int mx = 0, smx = 0;
  for (int i = head[p]; i; i = e[i].next) {
    int v = e[i].link;
    if (v != f) {
      dfs(v, p, NEKO);
      if (s[v] > mx) {
        smx = mx;
        mx = s[v];
      } else
        smx = max(smx, s[v]);
    }
  }
  if (mx + smx + 1 >= NEKO) {
    s[p] = 0;
    res++;
  } else
    s[p] = mx + 1;
}
inline void sol(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = (l + r) >> 1;
  res = 0;
  dfs(1, 0, mid);
  ans[mid] = res;
  sol(l, mid - 1, res, R);
  sol(mid + 1, r, L, res);
}
int main() {
  n = read();
  S = sqrt(n * log2(n));
  for (int i = 1, u, v; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  sol(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    write(ans[i]);
    pc('\n');
  }
  pc();
}
