Problem: Let G be a simple graph. Let W be a non-empty subset of vertices. Then W is almost-k-uniform if for each pair of distinct vertices u,v ∈ W the distance between u and v is either k or k+1.

You are given a tree on n vertices. For each i between 1 and n, find the maximum size of an almost-i-uniform set.

Input

The first line contains a single integer n (2 ≤ n ≤ 5 ⋅ 10^5) – the number of vertices of the tree.

Then n-1 lines follows, the i-th of which consisting of two space separated integers u_i, v_i (1 ≤ u_i, v_i ≤ n) meaning that there is an edge between vertices u_i and v_i. 

It is guaranteed that the given graph is tree. 

Output

Output a single line containing n space separated integers a_i, where a_i is the maximum size of an almost-i-uniform set.

Examples

Input


5
1 2
1 3
1 4
4 5


Output


4 3 2 1 1


Input


6
1 2
1 3
1 4
4 5
4 6


Output


4 4 2 1 1 1

Note

Consider the first example. 

  * The only maximum almost-1-uniform set is \{1, 2, 3, 4\}. 
  * One of the maximum almost-2-uniform sets is or \{2, 3, 5\}, another one is \{2, 3, 4\}. 
  * A maximum almost-3-uniform set is any pair of vertices on distance 3. 
  * Any single vertex is an almost-k-uniform set for k ≥ 1. 



In the second sample there is an almost-2-uniform set of size 4, and that is \{2, 3, 5, 6\}.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 500500;
int n, ans[N], distDown[N], distUp[N];
vector<int> a[N], dists[N];
void dfsDown(int x, int par) {
  for (int y : a[x])
    if (y != par) {
      dfsDown(y, x);
      dists[x].push_back(distDown[y] + 1);
      distDown[x] = max(distDown[x], distDown[y] + 1);
    }
}
void dfsUp(int x, int par) {
  int maxDist = distUp[x], maxChild = par, secondMaxDist = 0;
  for (int y : a[x])
    if (y != par) {
      if (distDown[y] + 1 > maxDist) {
        secondMaxDist = maxDist;
        maxDist = distDown[y] + 1;
        maxChild = y;
      } else
        secondMaxDist = max(secondMaxDist, distDown[y] + 1);
    }
  for (int y : a[x])
    if (y != par) {
      distUp[y] = (y == maxChild ? secondMaxDist : maxDist) + 1;
      dists[y].push_back(distUp[y]);
      dfsUp(y, x);
    }
}
void solve(int x, int par) {
  for (int y : a[x])
    if (y != par) {
      solve(y, x);
      for (int i = 1, j = 1; i < dists[x].size() && j < dists[y].size();) {
        int k = min(dists[x][i], dists[y][j]);
        ans[k * 2] = max(ans[k * 2], i + j);
        if (i + 1 == dists[x].size())
          j++;
        else if (j + 1 == dists[y].size())
          i++;
        else if (dists[x][i + 1] > dists[y][j + 1])
          i++;
        else
          j++;
      }
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfsDown(1, 0);
  dfsUp(1, 0);
  for (int i = 1; i <= n; i++) ans[i] = 1;
  for (int i = 1; i <= n; i++) {
    ans[1] = max(ans[1], int(a[i].size()) + 1);
    sort(dists[i].begin(), dists[i].end(), greater<int>());
    for (int j = 1; j < dists[i].size(); j++) {
      int k = min(dists[i][j - 1] - 1, dists[i][j]);
      ans[k * 2 + 1] = max(ans[k * 2 + 1], j + 1);
      ans[dists[i][j] * 2] = max(ans[dists[i][j] * 2], j + 1);
    }
  }
  solve(1, 0);
  for (int i = n - 2; i; i--) ans[i] = max(ans[i], ans[i + 2]);
  for (int i = 1; i <= n; i++) cout << ans[i] << " \n"[i == n];
}