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;
const int N = 100000 + 5;
const int BLOCK = 100;
int n;
int f[N][2];
vector<int> node;
vector<int> g[N];
int fa[N];
void dfs(int u, int f = 1) {
  for (auto v : g[u]) {
    if (v != f) {
      dfs(v, u);
    }
  }
  node.push_back(u);
  fa[u] = f;
}
int solve(int k) {
  for (int i = 1; i <= n; i++) {
    f[i][0] = f[i][1] = 0;
  }
  int ret = 0;
  for (int u : node) {
    if (f[u][0] + f[u][1] + 1 >= k) {
      ret++;
    } else {
      if (f[fa[u]][0] < f[u][0] + 1) {
        f[fa[u]][1] = f[fa[u]][0];
        f[fa[u]][0] = f[u][0] + 1;
      } else {
        f[fa[u]][1] = max(f[fa[u]][1], f[u][0] + 1);
      }
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  int x = N, k = 1;
  while (x > BLOCK) {
    x = solve(k++);
    printf("%d\n", x);
  }
  for (int i = BLOCK; i >= 0; i--) {
    if (solve(k) != i || k > n) {
      continue;
    }
    int l = k, r = n + 1;
    while (r - l > 1) {
      int mid = (l + r) / 2;
      if (solve(mid) == i) {
        l = mid;
      } else {
        r = mid;
      }
    }
    while (k <= l) {
      printf("%d\n", i);
      k++;
    }
  }
  return 0;
}
