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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 1e5 + 5;
int n, cnt;
int ans[N], f[N];
struct Edge {
  int to;
  Edge *nxt;
  Edge(int to = 0, Edge *nxt = NULL) : to(to), nxt(nxt) {}
} * head[N], pool[N << 1], *tail = pool;
inline void add(int u, int v) { head[u] = new (tail++) Edge(v, head[u]); }
void dfs(int x, int fa, int mid) {
  int mx1 = 0, mx2 = 0;
  for (Edge *i = head[x]; i; i = i->nxt) {
    int to = i->to;
    if (to == fa) continue;
    dfs(to, x, mid);
    if (f[to] > mx1)
      mx2 = mx1, mx1 = f[to];
    else if (f[to] > mx2)
      mx2 = f[to];
  }
  if (mx1 + mx2 + 1 >= mid)
    f[x] = 0, cnt++;
  else
    f[x] = mx1 + 1;
}
void solve(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = (l + r) >> 1;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  solve(l, mid - 1, cnt, R);
  solve(mid + 1, r, L, cnt);
}
int main() {
  n = read();
  for (int i = 1, u, v; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  fclose(stdin);
  fclose(stdout);
  return 0;
}
