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 = 100010;
vector<int> G[N];
int fa[N];
pair<int, int> dp[N];
vector<int> children[N];
int ans[N];
int n;
int k;
int rt;
void build() {
  rt = 1;
  queue<int> q;
  q.push(rt);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : G[u])
      if (fa[u] != v) {
        fa[v] = u;
        children[u].push_back(v);
        q.push(v);
      }
  }
}
void dfs(int r) {
  int a = 0;
  int b0 = 0;
  int b1 = 0;
  int b = 0;
  for (auto x : children[r]) {
    dfs(x);
    a += dp[x].first;
    if (dp[x].second > b0) {
      b1 = b0;
      b0 = dp[x].second;
    } else if (dp[x].second > b1) {
      b1 = dp[x].second;
    }
  }
  if (b0 + b1 + 1 >= k) {
    a++;
  } else {
    b = b0 + 1;
  }
  dp[r] = {a, b};
}
void getans(int l, int r) {
  if (ans[l] == ans[r]) {
    for (int i = l + 1; i <= r - 1; i++) {
      ans[i] = ans[l];
    }
    return;
  }
  if (r - l == 1) {
    return;
  }
  k = (r + l) >> 1;
  dfs(rt);
  ans[k] = dp[rt].first;
  getans(l, k);
  getans(k, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  build();
  for (int i = 2; i <= n; i++) {
    ans[i] = -1;
  }
  ans[1] = n;
  ans[n + 1] = 0;
  getans(1, n + 1);
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
}
