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 maxn = 1000000 + 5;
struct edge {
  int to, next;
} e[maxn];
int n, np, first[maxn], ans[maxn];
void addedge(int u, int v) {
  e[++np] = (edge){v, first[u]};
  first[u] = np;
}
pair<int, int> dp(int i, int f, int k) {
  pair<int, int> now = make_pair(0, 1);
  bool done = 0;
  for (int p = first[i]; p; p = e[p].next) {
    int j = e[p].to;
    if (j == f) continue;
    pair<int, int> x = dp(j, i, k);
    now.first += x.first;
    if (now.second + x.second >= k) done = 1;
    now.second = max(now.second, x.second + 1);
  }
  if (now.second >= k) done = 1;
  if (done)
    return make_pair(now.first + 1, 0);
  else
    return now;
}
void solve(int L, int R, int x, int y) {
  if (L > R || x < y) return;
  if (x == y) {
    for (int i = L; i <= R; i++) ans[i] = x;
    return;
  }
  int mid = (L + R) >> 1, cnt = dp(1, 1, mid).first;
  ans[mid] = cnt;
  solve(L, mid - 1, x, cnt);
  solve(mid + 1, R, cnt, y);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  solve(1, n, dp(1, 1, 1).first, dp(1, 1, n).first);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
