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>
#pragma GCC optimize("Ofast,unroll-loops")
using namespace std;
template <typename T>
void cmax(T& x, T y) {
  if (x < y) x = y;
}
template <typename T>
void cmin(T& x, T y) {
  if (x > y) x = y;
}
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const int N = 100005;
int n, al, num, h[N], dp[N], ans[N];
struct E {
  int x, t;
} e[2 * N];
inline void add(int s, int t) { e[++num].t = t, e[num].x = h[s], h[s] = num; }
inline void dfs(int u, int fa, int x) {
  int t1 = 0, t2 = 0;
  for (int i = h[u]; i; i = e[i].x)
    if (e[i].t ^ fa) {
      dfs(e[i].t, u, x);
      if (dp[e[i].t] >= t1)
        t2 = t1, t1 = dp[e[i].t];
      else if (dp[e[i].t] >= t2)
        t2 = dp[e[i].t];
    }
  if (t1 + t2 + 1 >= x)
    al++, dp[u] = 0;
  else
    dp[u] = t1 + 1;
}
inline void solve(int vl, int vr, int ql, int qr) {
  if (ql > qr) return;
  if (vl == vr) {
    for (int i = (ql), iend = (qr); i <= iend; i++) ans[i] = vl;
    return;
  }
  int mid = (ql + qr) >> 1;
  al = 0, dfs(1, 0, mid), ans[mid] = al;
  solve(vl, al, mid + 1, qr);
  solve(al, vr, ql, mid - 1);
}
inline void write(int x) {
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
int main() {
  n = read();
  for (int i = (2), iend = (n); i <= iend; i++) {
    int x = read(), y = read();
    add(x, y), add(y, x);
  }
  solve(0, n, 1, n);
  for (int i = (1), iend = (n); i <= iend; i++) write(ans[i]), puts("");
  return 0;
}
