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, point[200005], mx[100005], se[100005], head[100005], nxt[200005],
    ans[100005];
int tmp, tot, last;
void addedge(int x, int y) {
  point[++tot] = y, nxt[tot] = head[x], head[x] = tot;
  point[++tot] = x, nxt[tot] = head[y], head[y] = tot;
}
void dfs(int x, int fa, int k) {
  mx[x] = 1, se[x] = 1;
  for (int i = head[x]; i; i = nxt[i]) {
    int y = point[i];
    if (y == fa) continue;
    dfs(y, x, k);
    if (mx[x] < mx[y] + 1)
      se[x] = mx[x], mx[x] = mx[y] + 1;
    else if (se[x] < mx[y] + 1)
      se[x] = mx[y] + 1;
  }
  if (mx[x] + se[x] - 1 >= k) tmp++, mx[x] = 0;
}
int check(int k) {
  tmp = 0;
  dfs(1, 0, k);
  return tmp;
}
void solve(int l, int r, int ll, int rr) {
  if (l > r) return;
  if (ll == rr) {
    for (int i = l; i <= r; i++) ans[i] = ll;
    return;
  }
  int mid = (l + r) >> 1;
  if (ans[mid] == -1) ans[mid] = check(mid);
  solve(l, mid - 1, ans[mid], rr);
  solve(mid + 1, r, ll, ans[mid]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    addedge(x, y);
  }
  memset(ans, -1, sizeof(ans));
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
