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 inf = 0x7fffffff;
const double eps = 1e-10;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch;
  ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch & 15);
    ch = getchar();
  }
  if (f)
    return x;
  else
    return -x;
}
const int N = 100010;
int head[N], p, ans[N];
struct ss {
  int v, last;
} G[N << 1];
void add(int a, int b) {
  G[++p] = (ss){b, head[a]};
  head[a] = p;
  G[++p] = (ss){a, head[b]};
  head[b] = p;
}
int n;
int Max[N], Cmax[N];
void C_max(int i, int c) {
  if (c >= Max[i])
    Cmax[i] = Max[i], Max[i] = c;
  else if (c > Cmax[i])
    Cmax[i] = c;
}
int sta[N], dfn[N], fa[N], tim;
void dfs(int a, int f) {
  sta[dfn[a] = ++tim] = a;
  fa[a] = f;
  for (int i = head[a]; i; i = G[i].last)
    if (G[i].v != f) dfs(G[i].v, a);
}
int dep[N];
int solve(int k) {
  int ret = 0;
  memset(Max, 0, sizeof(Max));
  memset(Cmax, 0, sizeof(Cmax));
  for (int i = n, u, v; i >= 1; i--) {
    u = sta[i];
    v = fa[u];
    if (!Max[u]) dep[u] = 1;
    if (Cmax[u] + Max[u] + 1 >= k)
      ++ret, dep[u] = 0;
    else
      dep[u] = Max[u] + 1;
    C_max(v, dep[u]);
  }
  return ret;
}
int find(int l, int r, int val) {
  while (l < r - 1) {
    int mid = l + r >> 1;
    if (ans[mid] == -1) ans[mid] = solve(mid);
    if (ans[mid] >= val)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int main() {
  memset(ans, -1, sizeof(ans));
  n = read();
  for (int i = 1, a, b; i < n; i++) a = read(), b = read(), add(a, b);
  dfs(1, 0);
  ans[1] = n;
  int size = sqrt(n);
  for (int i = 2; i <= size; i++) ans[i] = solve(i);
  for (int l = size + 1, r; l <= n; l = r + 1) {
    if (ans[l] == -1) ans[l] = solve(l);
    r = find(l, n + 1, ans[l]);
    for (int i = l + 1; i <= r; i++) ans[i] = ans[l];
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
