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 N = 1e5 + 3;
struct Lst {
  int dir, nxt;
} E[N << 1];
int G[N], cnt;
inline void Add(int u, int v) {
  E[++cnt].dir = v, E[cnt].nxt = G[u], G[u] = cnt;
}
int n, dfn[N], fa[N], ord;
void dfs(int u) {
  for (register int i = G[u]; i; i = E[i].nxt) {
    int v = E[i].dir;
    if (v == fa[u]) continue;
    fa[v] = u, dfs(v);
  }
  dfn[++ord] = u;
}
inline int solve(int k) {
  static int dp[N], res;
  for (register int i = (1); i <= (n); ++i) dp[i] = 1;
  res = 0;
  for (register int i = (1); i <= (n); ++i) {
    static int u;
    u = dfn[i];
    if (fa[u] && (~dp[fa[u]]) && (~dp[u]))
      if (dp[fa[u]] + dp[u] >= k)
        dp[fa[u]] = -1, res++;
      else
        dp[fa[u]] = max(dp[fa[u]], dp[u] + 1);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (register int i = (2); i <= (n); ++i) {
    static int u, v;
    scanf("%d%d", &u, &v);
    Add(u, v), Add(v, u);
  }
  dfs(1);
  static int lim;
  lim = sqrt(n), printf("%d\n", n);
  for (register int i = (2); i <= (lim); ++i) printf("%d\n", solve(i));
  for (register int l = lim + 1, r, std; l <= n; l = r + 1) {
    static int L, R, M;
    r = L = l, R = n, std = solve(l);
    while (L <= R) {
      M = L + R >> 1;
      if (solve(M) >= std)
        r = M, L = M + 1;
      else
        R = M - 1;
    }
    for (register int i = (l); i <= (r); ++i) printf("%d\n", std);
  }
  return 0;
}
