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() {
  int x = 0, ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
inline void write(int x) {
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 1e5 + 5;
struct edge {
  int link, next;
} e[N << 1];
int n, tot, head[N];
inline void add_edge(int u, int v) {
  e[++tot] = (edge){v, head[u]}, head[u] = tot;
}
int dp[N], ans[N], rd;
void dfs(int u, int fa, int x) {
  dp[u] = 0;
  int mx1, mx2;
  mx1 = mx2 = 0;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].link;
    if (v != fa) {
      dfs(v, u, x);
      if (dp[v] > mx1)
        mx2 = mx1, mx1 = dp[v];
      else
        mx2 = max(mx2, dp[v]);
    }
  }
  if (mx1 + mx2 + 1 >= x)
    dp[u] = 0, rd++;
  else
    dp[u] = mx1 + 1;
}
void work(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;
  rd = 0;
  dfs(1, -1, mid);
  ans[mid] = rd;
  work(l, mid - 1, rd, R);
  work(mid + 1, r, L, rd);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    add_edge(u, v);
    add_edge(v, u);
  }
  work(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    write(ans[i]);
    putchar('\n');
  }
  return 0;
}
