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 = 1e6 + 5;
int n, m, gs, top, f[N], Ans[N], head[N];
vector<int> g[N];
struct E {
  int too, Nxt;
} edge[N * 2];
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline int gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  int ret = 0, f = 0;
  char c = gc();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = gc();
  }
  while (isdigit(c)) {
    ret = ret * 10 + c - 48;
    c = gc();
  }
  if (f) return -ret;
  return ret;
}
void add(int a, int b) {
  edge[++top].too = b;
  edge[top].Nxt = head[a];
  head[a] = top;
}
void write(int x) {
  if (x <= 9) {
    putchar(x + '0');
    return;
  }
  write(x / 10);
  putchar(x % 10 + '0');
}
void dfs(int u, int fa, int mid) {
  f[u] = 0;
  int zd = 0, cd = 0;
  for (int i = head[u]; i; i = edge[i].Nxt) {
    int v = edge[i].too;
    if (v == fa) continue;
    dfs(v, u, mid);
    if (f[v] > zd) {
      cd = zd;
      zd = f[v];
    } else if (f[v] > cd)
      cd = f[v];
  }
  if (zd + cd + 1 >= mid)
    gs++;
  else
    f[u] = zd + 1;
}
void solve(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) Ans[i] = L;
    return;
  }
  int mid = (l + r) / 2;
  gs = 0;
  dfs(1, 0, mid);
  Ans[mid] = gs;
  solve(l, mid - 1, gs, R);
  solve(mid + 1, r, L, gs);
}
signed main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    write(Ans[i]);
    putchar('\n');
  }
  return 0;
}
