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;
inline int read() {
  char ch = getchar();
  int w = 1, s = 0;
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  };
  while (isdigit(ch)) {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return w * s;
}
struct node {
  int to, next;
} e[1000010];
int n, h[1000010], tot;
inline void add(int from, int to) {
  e[++tot].next = h[from];
  h[from] = tot;
  e[tot].to = to;
}
int ANS[1000100], dp[1000010], len[1000010];
inline void DFS(int now, int ffa, int ud) {
  int opt = 0;
  int Sum = 0;
  int Max1 = 0, Max2 = 0;
  for (register int i = h[now]; i; i = e[i].next) {
    int to = e[i].to;
    if (to == ffa) continue;
    DFS(to, now, ud);
    opt = 1;
    Sum += dp[to];
    if (len[to] > Max1)
      Max2 = max(Max2, Max1), Max1 = max(Max1, len[to]);
    else
      Max2 = max(Max2, len[to]);
  }
  if (!opt) {
    dp[now] = 0;
    len[now] = 1;
    return;
  } else {
    if (!Max2) {
      if (Max1 + 1 >= ud) {
        dp[now] = Sum + 1;
        len[now] = Max1 + 1 - ud;
      } else
        dp[now] = Sum, len[now] = Max1 + 1;
    } else {
      if (Max2 + Max1 + 1 >= ud) {
        dp[now] = Sum + 1;
        len[now] = max(0, Max1 + 1 - ud);
      } else {
        dp[now] = Sum;
        len[now] = Max1 + 1;
      }
    }
  }
}
inline int calc(int now) {
  DFS(1, 0, now);
  return dp[1];
}
int main() {
  n = read();
  for (register int i = 1; i < n; ++i) {
    int u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  ANS[1] = n;
  int tmp = 1300;
  for (register int i = 2; i <= min(n, 1300); ++i) {
    ANS[i] = calc(i);
  }
  while (tmp <= n) {
    int l = tmp, r = n, ans;
    int cur = calc(tmp);
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (n / mid < cur) {
        r = mid - 1;
        continue;
      }
      if (calc(mid) == cur) {
        l = mid + 1, ans = mid;
      } else
        r = mid - 1;
    }
    for (register int i = tmp; i <= ans; ++i) {
      ANS[i] = cur;
    }
    tmp = ans + 1;
  }
  for (register int i = 1; i <= n; ++i) cout << ANS[i] << "\n";
  return 0;
}
