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 long long read() {
  long long 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 << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 100008;
int n;
int x, y;
int nume, head[N];
struct edge {
  int to, nxt;
} e[N << 1];
inline void add_edge(int x, int y) {
  e[++nume] = (edge){y, head[x]};
  head[x] = nume;
}
int ans[N];
int tmp;
int dp[N];
inline void dfs(int x, int fa, int goal) {
  int mx1, mx2;
  mx1 = mx2 = 0;
  for (int i = head[x]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to != fa) {
      dfs(to, x, goal);
      if (dp[to] > mx1) {
        mx2 = mx1;
        mx1 = dp[to];
      } else if (dp[to] > mx2) {
        mx2 = dp[to];
      }
    }
  }
  if (mx1 + mx2 + 1 >= goal) {
    dp[x] = 0;
    ++tmp;
  } else {
    dp[x] = mx1 + 1;
  }
}
inline void solve(int l, int r, int ansl, int ansr) {
  if (l > r) return;
  if (ansl == ansr) {
    for (int i = l; i <= r; ++i) {
      ans[i] = ansl;
    }
    return;
  }
  int mid = (l + r) >> 1;
  tmp = 0;
  dfs(1, 0, mid);
  ans[mid] = tmp;
  solve(l, mid - 1, ansl, ans[mid]);
  solve(mid + 1, r, ans[mid], ansr);
}
int main() {
  n = read();
  for (int i = 1; i < n; ++i) {
    x = read();
    y = read();
    add_edge(x, y);
    add_edge(y, x);
  }
  solve(1, n, n, 0);
  for (int i = 1; i <= n; ++i) {
    printf("%d ", ans[i]);
  }
  return 0;
}
