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>
#pragma GCC optimize "-O3"
using namespace std;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') {
      f = -1;
    }
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 100100;
vector<int> ver[N], son[N], v;
int path[N], ans[N];
void add(int x, int y) {
  ver[x].push_back(y);
  ver[y].push_back(x);
}
void dfs(int x, int fa) {
  for (auto y : ver[x]) {
    if (y != fa) {
      son[x].push_back(y);
      dfs(y, x);
    }
  }
  v.push_back(x);
}
int calc(int k) {
  int ret = 0;
  for (auto x : v) {
    int mx1 = 0, mx2 = 0;
    for (auto y : son[x]) {
      if (path[y] > mx1) {
        mx2 = mx1;
        mx1 = path[y];
      } else if (path[y] > mx2) {
        mx2 = path[y];
      }
    }
    if (mx1 + mx2 + 1 >= k) {
      ret++;
      path[x] = 0;
    } else {
      path[x] = mx1 + 1;
    }
  }
  return ret;
}
void solve(int l, int r, int al, int ar) {
  if (al == ar) {
    for (int i = l; i <= r; i++) {
      ans[i] = al;
    }
    return;
  }
  int mid = l + r >> 1;
  solve(l, mid, al, calc(mid));
  solve(mid + 1, r, calc(mid + 1), ar);
}
int main() {
  int n = read();
  for (int i = 1; i < n; i++) {
    add(read(), read());
  }
  dfs(1, 0);
  solve(1, n, calc(1), calc(n));
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
