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;
const int MAXN = 2e5 + 5;
template <typename T>
void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
vector<int> a[MAXN];
int n, tot, father[MAXN], p[MAXN], ans[MAXN];
void work(int pos, int fa) {
  father[pos] = fa;
  for (unsigned i = 0; i < a[pos].size(); i++)
    if (a[pos][i] != fa) work(a[pos][i], pos);
  p[++tot] = pos;
}
int f(int x) {
  if (ans[x] != -1) return ans[x];
  static int res[MAXN];
  int tans = 0;
  for (int i = 1; i <= n; i++) {
    int Max = 1, Nax = 1, pos = p[i];
    for (unsigned i = 0; i < a[pos].size(); i++)
      if (a[pos][i] != father[pos]) {
        int tmp = res[a[pos][i]] + 1;
        if (tmp > Max) {
          Nax = Max;
          Max = tmp;
        } else
          chkmax(Nax, tmp);
      }
    if (Max + Nax - 1 >= x)
      tans++, res[pos] = 0;
    else
      res[pos] = Max;
  }
  return ans[x] = tans;
}
int main() {
  read(n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x), read(y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  work(1, 0);
  memset(ans, -1, sizeof(ans));
  for (int i = 1, nxt; i <= n; i = nxt + 1) {
    int now = f(i);
    if (now >= 65 || f(i + 1) != now) {
      ans[nxt = i] = now;
      continue;
    }
    int l = i + 1, r = n;
    while (l < r) {
      int mid = (l + r + 1) / 2;
      if (f(mid) == now)
        l = mid;
      else
        r = mid - 1;
    }
    nxt = l;
    for (int j = i; j <= nxt; j++) ans[j] = now;
  }
  for (int i = 1; i <= n; i++) writeln(ans[i]);
  return 0;
}
