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 = 1e5 + 5;
using namespace std;
struct node {
  int x, y;
} q[N << 1];
int k[N], f[N], da[N], df[N], fa[N];
int ans = 0, n, su, s;
inline int r() {
  int y = 0;
  char h = getchar();
  while (h < '0' || h > '9') h = getchar();
  while (h >= '0' && h <= '9') y = y * 10 + h - '0', h = getchar();
  return y;
}
inline void jr(int a, int b) {
  q[++su] = (node){k[a], b};
  k[a] = su;
}
void dfs(int x, int f) {
  for (int i = k[x]; i; i = q[i].x) {
    int y = q[i].y;
    if (y == f) continue;
    fa[y] = x;
    dfs(y, x);
  }
  df[++s] = x;
}
inline int work(int k) {
  if (da[k] != -1) return da[k];
  for (int i = 1; i <= n; i++) f[i] = 1;
  ans = 0;
  for (int i = 1; i <= n; i++) {
    int y = df[i], x = fa[y];
    if (!x) continue;
    if (f[x] != -1 && f[y] != -1) {
      if (f[x] + f[y] >= k)
        ans++, f[x] = f[y] = -1;
      else
        f[x] = max(f[x], f[y] + 1);
    }
  }
  return da[k] = ans;
}
signed main() {
  for (int i = 0; i < N; i++) da[i] = -1;
  n = r();
  int p = sqrt(n * log2(n));
  for (int i = 1; i < n; i++) {
    int x = r(), y = r();
    jr(x, y);
    jr(y, x);
  }
  dfs(1, 0);
  printf("%d\n", n);
  for (int i = 2; i <= p; i++) printf("%d\n", work(i));
  for (int i = p + 1; i <= n;) {
    int x = work(i);
    int l = i, r = n, daa = 0;
    while (l <= r) {
      int m = (l + r) >> 1;
      if (work(m) == x)
        daa = m, l = m + 1;
      else
        r = m - 1;
    }
    for (int j = i; j <= daa; j++) da[j] = x;
    i = daa + 1;
  }
  for (int i = p + 1; i <= n; i++) printf("%d\n", work(i));
}
