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 rd() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const int maxn = 1e5 + 233;
struct Edge {
  int to, nxt;
  Edge() {}
  Edge(int to, int nxt) : to(to), nxt(nxt) {}
} edge[maxn * 2];
int first[maxn], nume;
int dp[maxn];
int V;
int res = 0;
inline void dfs(int u, int fa) {
  int mx, mx2;
  mx = mx2 = 0;
  for (int e = first[u]; ~e; e = edge[e].nxt) {
    int v = edge[e].to;
    if (v == fa) continue;
    dfs(v, u);
    if (dp[v] > mx)
      mx2 = mx, mx = dp[v];
    else {
      mx2 = max(mx2, dp[v]);
    }
  }
  if (mx + mx2 + 1 >= V) {
    dp[u] = 0;
    res++;
  } else
    dp[u] = mx + 1;
}
inline int get(int x) {
  V = x;
  res = 0;
  dfs(1, 0);
  return res;
}
int ans[maxn];
inline void solve(int l, int r, int L, int R) {
  if (L == R) {
    for (register int i = (l); i <= int(r); ++i) ans[i] = L;
    return;
  }
  int mid = (l + r) >> 1;
  int w = get(mid);
  ans[mid] = w;
  if (l <= mid - 1) solve(l, mid - 1, w, R);
  if (mid + 1 <= r) solve(mid + 1, r, L, w);
}
int n;
void addedge(int a, int b) {
  edge[nume] = Edge(b, first[a]);
  first[a] = nume++;
}
int main() {
  memset(first, -1, sizeof(first));
  nume = 0;
  n = rd();
  for (register int i = (1); i < int(n); ++i) {
    int a = rd(), b = rd();
    addedge(a, b);
    addedge(b, a);
  }
  solve(1, n, 0, n);
  for (register int i = (1); i <= int(n); ++i) writeln(ans[i]);
  return 0;
}
