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;
int n;
int fa[100005], ans[100005], f[100005];
vector<int> e[100005], v;
const int B = 200;
inline void addEdge(int u, int v) { e[u].push_back(v); }
void dfs(int u, int faa) {
  for (int v : e[u])
    if (v != faa) dfs(v, u);
  fa[u] = faa, v.push_back(u);
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
    if (ch == -1) return 0;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline int solve(int x) {
  if (ans[x]) return ans[x];
  int res = 0;
  for (int i = 1; i <= n; i++) f[i] = 1;
  for (int i = 1; i <= n; i++) {
    int u = v[i];
    if (fa[u] && (~f[u]) && (~f[fa[u]]))
      if (f[u] + f[fa[u]] >= x)
        res++, f[fa[u]] = -1;
      else
        f[fa[u]] = max(f[fa[u]], f[u] + 1);
  }
  return ans[x] = res;
}
int main() {
  n = read();
  for (int i = 1, u, v; i < n; i++)
    u = read(), v = read(), addEdge(u, v), addEdge(v, u);
  v.push_back(0), dfs(1, 0), ans[1] = n;
  for (int i = 2; i <= min(n, B); i++) ans[i] = solve(i);
  for (int i = min(n, B) + 1, j = 0; i <= n; i = j + 1) {
    int x = solve(i), l = i, r = n;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (solve(mid) ^ x)
        r = mid - 1;
      else
        j = mid, l = mid + 1;
    }
    for (int k = i; k <= j; k++) ans[k] = x;
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
