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 = 1e5 + 10;
int n, cnt_edge, nowans;
int head[maxn], f[maxn], ans[maxn];
struct edge {
  int to, nxt;
} e[maxn << 1];
inline int read() {
  char c = getchar();
  int res = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res * f;
}
inline void add_edge(int u, int v) {
  e[++cnt_edge].nxt = head[u];
  head[u] = cnt_edge;
  e[cnt_edge].to = v;
}
void dfs(int x, int fa, int m) {
  f[x] = 0;
  int maxx1 = 0, maxx2 = 0;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs(y, x, m);
    if (f[y] > maxx1)
      maxx2 = maxx1, maxx1 = f[y];
    else if (f[y] > maxx2)
      maxx2 = f[y];
  }
  if (maxx1 + maxx2 + 1 >= m)
    nowans++, f[x] = 0;
  else
    f[x] = maxx1 + 1;
}
void solve(int ql, int qr, int l, int r) {
  if (ql > qr) return;
  if (l == r) {
    for (int i = ql; i <= qr; i++) ans[i] = l;
    return;
  }
  int mid = (ql + qr) >> 1;
  nowans = 0;
  dfs(1, 0, mid);
  ans[mid] = nowans;
  solve(ql, mid - 1, ans[mid], r);
  solve(mid + 1, qr, l, ans[mid]);
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    add_edge(u, v), add_edge(v, u);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
