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;
vector<int> adjList[100000];
pair<int, int> dp[100000];
int num[100000], c = 0;
vector<int> adjList2[100000];
int doDFS(int u, int p) {
  int i;
  num[u] = c++;
  for (i = 0; i < adjList[u].size(); i++) {
    int v = adjList[u][i];
    if (v != p) doDFS(v, u), adjList2[num[u]].push_back(num[v]);
  }
  return 0;
}
int ans[100001];
int getAns(int n, int k) {
  if (ans[k] != -1) return ans[k];
  int u, i;
  for (u = n - 1; u >= 0; u--) {
    int paths = 0, len1 = 0, len2 = 0;
    for (i = 0; i < adjList2[u].size(); i++) {
      int v = adjList2[u][i];
      paths += dp[v].first;
      if (dp[v].second >= len1)
        len2 = len1, len1 = dp[v].second;
      else if (dp[v].second >= len2)
        len2 = dp[v].second;
    }
    if (len1 + len2 + 1 >= k)
      dp[u] = make_pair(paths + 1, 0);
    else
      dp[u] = make_pair(paths, len1 + 1);
  }
  return ans[k] = dp[0].first;
}
int main() {
  int i;
  int n, u, v;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    u--, v--;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  doDFS(0, -1);
  fill(ans, ans + n + 1, -1);
  int B = sqrt(n);
  for (i = 1; i <= B; i++) ans[i] = getAns(n, i);
  int c = B;
  while (c < n) {
    int l = c, r = n;
    while (l < r) {
      int m = (l + r + 1) / 2;
      if (getAns(n, m) == ans[c])
        l = m;
      else
        r = m - 1;
    }
    fill(ans + c, ans + l + 1, ans[c]);
    c = l + 1, ans[c] = getAns(n, l + 1);
  }
  for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
