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>
const int maxn = 100000 + 10, maxm = 200000 + 10, sq = 500;
int n, pre, su[maxn], le[maxn], ans[maxn], ma[maxn], seg[maxn], max1[maxn],
    max2[maxn], top;
std::vector<int> e[maxn];
void dfs(int u, int fa) {
  ma[u] = fa;
  for (int i = e[u].size() - 1; i >= 0; --i) {
    int v = e[u][i];
    if (v == fa) continue;
    dfs(v, u);
  }
  seg[++top] = u;
}
int calc(int k) {
  memset(su, 0, sizeof(su));
  memset(le, 0, sizeof(le));
  memset(max1, 0, sizeof(max1));
  memset(max2, 0, sizeof(max2));
  int u, v;
  for (int i = 1; i <= n; ++i) {
    v = seg[i];
    u = ma[v];
    if (max1[v] + max2[v] >= k)
      ++su[v];
    else
      le[v] = max1[v] + 1;
    su[u] += su[v];
    if (le[v] > max1[u]) {
      max2[u] = max1[u];
      max1[u] = le[v];
    } else if (le[v] > max2[u])
      max2[u] = le[v];
  }
  return su[1];
}
void meow(int x) {
  for (int l = sq + 1, r = pre, mid; l <= r;) {
    mid = (l + r) / 2;
    if (calc(mid) >= x) {
      ans[mid] = x;
      l = mid + 1;
    } else
      pre = r = mid - 1;
  }
}
int read() {
  int ret = 0;
  char ch = getchar();
  while ('9' < ch || ch < '0') ch = getchar();
  while ('0' <= ch && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret;
}
int main() {
  int u, v;
  n = read();
  pre = n - 1;
  for (int i = 1; i < n; ++i) {
    u = read();
    v = read();
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 0; i < n && i <= sq; ++i) ans[i] = calc(i);
  for (int i = 1; i <= n / sq; ++i) meow(i);
  for (int i = n - 2; i >= 0; --i)
    ans[i] = ((ans[i]) > (ans[i + 1]) ? (ans[i]) : (ans[i + 1]));
  for (int i = 0; i < n; ++i) printf("%d\n", ans[i]);
  return 0;
}
