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;
long long read() {
  long long x = 0, f = 1;
  char ch = ' ';
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') f = -1;
  }
  while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
int n, ans, tot, h[100005];
vector<int> g[100005];
struct AB {
  int a, b, n;
} d[100005];
void charu(int a, int b) {
  d[++tot].a = a, d[tot].b = b, d[tot].n = h[a], h[a] = tot;
}
int dfs(int u, int num) {
  int first = 0, second = 0;
  for (int k = h[u]; k; k = d[k].n) {
    int v = d[k].b;
    int tmp = dfs(v, num);
    if (tmp > first)
      second = first, first = tmp;
    else if (tmp > second)
      second = tmp;
  }
  if (first + second + 1 >= num) {
    ans++;
    return 0;
  }
  return first + 1;
}
void dfs1(int u, int fa) {
  for (int v : g[u]) {
    if (v == fa) continue;
    charu(u, v);
    dfs1(v, u);
  }
}
int solve(int num) {
  ans = 0;
  dfs(1, num);
  return ans;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a = read(), b = read();
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(1, 0);
  int p = sqrt(n * (log(n) / log(2)));
  ;
  cout << n << endl;
  for (int i = 2; i <= p; i++) {
    printf("%d\n", solve(i));
  }
  int di = solve(p), las = p;
  for (int i = di; i >= 0; i--) {
    if (solve(las + 1) != i) continue;
    int l = las + 1, r = n, mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (solve(mid) == i)
        l = mid;
      else
        r = mid - 1;
    }
    for (int j = las + 1; j <= r; j++) printf("%d\n", i);
    las = r;
  }
}
