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 NMax = 100005, limit = 1303;
int N, K, Res[NMax], last;
int Nb[NMax];
pair<int, int> DP[NMax];
vector<int> G[NMax];
int Father[NMax], Ord[NMax], cnt, Max[NMax], Max2[NMax], Sum[NMax];
void Read() {
  scanf("%d", &N);
  Nb[N] = -1;
  for (int i = 1; i < N; i++) {
    int x, y;
    Nb[i] = -1;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
}
void DFS(int node, int father) {
  Father[node] = father;
  Ord[++cnt] = node;
  for (int i = 0; i < G[node].size(); i++) {
    int neighb = G[node][i];
    if (neighb == father) continue;
    DFS(neighb, node);
  }
}
void Bernabeu(int K) {
  for (int i = 1; i <= N; i++) Max[i] = Max2[i] = Sum[i] = 0;
  for (int i = N; i >= 1; i--) {
    int node = Ord[i];
    int father = Father[node];
    if (Max[node] + Max2[node] + 1 >= K)
      DP[node] = make_pair(0, Sum[node] + 1);
    else
      DP[node] = make_pair(Max[node] + 1, Sum[node]);
    Sum[father] += DP[node].second;
    if (Max[father] < DP[node].first) {
      Max2[father] = Max[father];
      Max[father] = DP[node].first;
    } else if (Max2[father] < DP[node].first)
      Max2[father] = DP[node].first;
  }
}
int findRes(int k) {
  if (Nb[k] != -1) return Nb[k];
  Bernabeu(k);
  Nb[k] = DP[1].second;
  return DP[1].second;
}
int binSearch1(int ans, int left, int right) {
  int mid, sol = limit;
  while (left <= right) {
    mid = (left + right) / 2;
    if (findRes(mid) > ans) {
      left = mid + 1;
      sol = mid;
    } else
      right = mid - 1;
  }
  return sol;
}
int binSearch2(int ans, int left, int right) {
  int mid, sol = N + 1;
  while (left <= right) {
    mid = (left + right) / 2;
    if (findRes(mid) >= ans) {
      left = mid + 1;
    } else {
      right = mid - 1;
      sol = mid;
    }
  }
  return sol;
}
void Solve(int ans) {
  int r = binSearch2(ans, limit + 1, N);
  for (int i = last + 1; i < r; i++) Res[i] = ans;
  last = r - 1;
}
void getRes() {
  for (int i = 1; i <= min(N, limit); i++) Res[i] = findRes(i);
  last = limit;
  for (int i = N / limit; i >= 1; i--) {
    Solve(i);
  }
  for (int i = 1; i <= N; i++) printf("%d\n", Res[i]);
}
int main() {
  Read();
  DFS(1, 0);
  getRes();
  return 0;
}
