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 + 5;
int n;
struct Edge {
  int to, nxt;
} e[2 * maxn];
int cnt;
int head[maxn];
int read() {
  int x = 0, y = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') y = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * y;
}
void add(int u, int v) {
  e[cnt] = (Edge){v, head[u]};
  head[u] = cnt++;
  return;
}
int dfsnum;
int dfn[maxn], rnk[maxn];
void dfs1(int x, int fa) {
  dfn[x] = ++dfsnum;
  rnk[dfsnum] = x;
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int tmp = e[i].to;
    if (tmp == fa) continue;
    dfs1(tmp, x);
  }
  return;
}
int k, s;
int mx[maxn];
void calc() {
  for (int i = n; i >= 1; i--) {
    int x = rnk[i];
    int mx1 = 0, mx2 = 0;
    for (int j = head[x]; j != -1; j = e[j].nxt) {
      int tmp = e[j].to;
      if (dfn[tmp] < i) continue;
      if (mx1 < mx[tmp])
        mx2 = mx1, mx1 = mx[tmp];
      else
        mx2 = max(mx2, mx[tmp]);
    }
    if (mx1 + mx2 + 1 >= k)
      s++, mx[x] = 0;
    else
      mx[x] = mx1 + 1;
  }
  return;
}
int main() {
  n = read();
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; i++) {
    int u, v;
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  dfs1(1, 0);
  int B = min(sqrt(n * 17), 1.0 * n), lst = n;
  for (int i = 1; i <= B; i++) {
    k = i, s = 0;
    calc();
    printf("%d\n", s);
    lst = s;
  }
  int pre = B;
  int l = B, r = n + 1, mid;
  for (int i = lst; i >= 1; i--) {
    while (l + 1 < r) {
      mid = l + ((r - l) >> 1);
      k = mid, s = 0;
      calc();
      if (s < i)
        r = mid;
      else
        l = mid;
    }
    for (int j = pre + 1; j <= l; j++) printf("%d\n", i);
    pre = l;
    r = n + 1;
  }
  for (int i = pre + 1; i <= n; i++) puts("0");
  return 0;
}
