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;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
const int N = 1e5 + 10;
struct ed {
  int to, next;
} e[N << 1];
int m, n, head[N], ecnt;
void ad(int x, int y) {
  e[++ecnt] = (ed){y, head[x]};
  head[x] = ecnt;
  e[++ecnt] = (ed){x, head[y]};
  head[y] = ecnt;
}
int ans;
int dfs(int u, int last) {
  int mx1 = 0, mx2 = 0;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    int tmp = dfs(v, u);
    if (tmp > mx1)
      mx2 = mx1, mx1 = tmp;
    else if (tmp > mx2)
      mx2 = tmp;
  }
  if (mx1 + mx2 + 1 >= m) {
    ans++;
    return 0;
  }
  return mx1 + 1;
}
int Ans[N];
int cal(int x) {
  if (Ans[x] != -1) return Ans[x];
  m = x, ans = 0;
  dfs(1, 0);
  return Ans[x] = ans;
}
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) printf("%d\n", L);
    return;
  }
  int mid = l + r >> 1, tmp = cal(mid);
  solve(l, mid - 1, tmp, R), printf("%d\n", tmp), solve(mid + 1, r, L, tmp);
}
int main() {
  memset(Ans, -1, sizeof(Ans));
  int x, y;
  read(n);
  for (int i = 1; i < n; ++i) read(x), read(y), ad(x, y);
  solve(1, n, 0, n);
  return 0;
}
