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 <typename T>
inline void read(T& x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
  x *= f;
}
struct node {
  int v, next;
} e[400050 << 1];
int n, x, y, tot, head[100050], ans[100050], f[100050], cnt;
void add(int u, int v) {
  e[++tot] = (node){v, head[u]};
  head[u] = tot;
}
void dfs(int u, int fa, int x) {
  int mx1 = 0, mx2 = 0;
  f[u] = 0;
  for (int i = head[u], v; i; i = e[i].next)
    if ((v = e[i].v) != fa) {
      dfs(v, u, x);
      if (f[v] > mx1)
        mx2 = mx1, mx1 = f[v];
      else
        mx2 = max(mx2, f[v]);
    }
  if (mx1 + mx2 + 1 >= x)
    f[u] = 0, cnt++;
  else
    f[u] = mx1 + 1;
}
void solve(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;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  solve(l, mid - 1, cnt, R);
  solve(mid + 1, r, L, cnt);
}
int main() {
  read(n);
  for (int i = 1; i < n; ++i) read(x), read(y), add(x, y), add(y, x);
  solve(1, n, 0, n);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
