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")
using namespace std;
const int MAXN = 1e5 + 51;
struct Edge {
  int to, prev;
};
Edge ed[MAXN << 1];
int nc, tot, p, from, to;
int last[MAXN], fa[MAXN], dp[MAXN], res[MAXN];
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
template <class T>
inline T Max(T x, T y) {
  return x > y ? x : y;
}
inline void addEdge(int from, int to) {
  ed[++tot].prev = last[from];
  ed[tot].to = to;
  last[from] = tot;
}
inline void dfs(int node, int f, int x) {
  fa[node] = f;
  int maxn = 0, maxn2 = 0;
  for (register int i = last[node]; i; i = ed[i].prev) {
    if (ed[i].to != f) {
      dfs(ed[i].to, node, x);
      if (dp[ed[i].to] > maxn) {
        maxn2 = maxn, maxn = dp[ed[i].to];
      } else {
        maxn2 = Max(maxn2, dp[ed[i].to]);
      }
    }
  }
  if (maxn + maxn2 + 1 >= x) {
    dp[node] = 0, p++;
  } else {
    dp[node] = maxn + 1;
  }
}
inline void solve(int l, int r, int lx, int rx) {
  if (l > r || lx > rx) {
    return;
  }
  if (lx == rx) {
    for (register int i = l; i <= r; i++) {
      res[i] = lx;
    }
    return;
  }
  int mid = (l + r) >> 1;
  p = 0, dfs(1, 0, mid), res[mid] = p;
  solve(l, mid - 1, p, rx), solve(mid + 1, r, lx, p);
}
int main() {
  nc = read();
  for (register int i = 1; i < nc; i++) {
    from = read(), to = read();
    addEdge(from, to), addEdge(to, from);
  }
  solve(1, nc, 0, nc);
  for (register int i = 1; i <= nc; i++) {
    printf("%d\n", res[i]);
  }
}
