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 long double eps = 1e-13;
const long double PI = acos(-1);
const int INF = (int)1e9;
const long long INFF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MXN = (int)2e5 + 7;
vector<int> edge[MXN], E[MXN];
int ans[MXN];
int dp[MXN][2];
int in[MXN], cntp;
void dfs(int u, int pa = 0) {
  for (int v : edge[u]) {
    if (v == pa) continue;
    dfs(v, u);
    E[u].emplace_back(v);
  }
  in[++cntp] = u;
}
int n;
int sol(int need) {
  for (int i = 1; i <= n; i++) {
    int u = in[i];
    dp[u][0] = dp[u][1] = 0;
    int ok = 0;
    for (int v : E[u]) {
      dp[u][0] += dp[v][0];
      if (dp[v][1] + 1 + dp[u][1] >= need) {
        ok = 1;
      }
      dp[u][1] = max(dp[u][1], dp[v][1]);
    }
    if (ok || dp[u][1] + 1 >= need) {
      dp[u][0]++;
      dp[u][1] = 0;
    } else {
      dp[u][1]++;
    }
  }
  return dp[1][0];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    edge[u].emplace_back(v);
    edge[v].emplace_back(u);
  }
  dfs(1);
  int h = min(1007, n);
  for (int i = 1; i <= h; i++) ans[i] = sol(i);
  int pos = h + 1;
  for (int i = ans[h]; i > 0; i--) {
    int ll = pos, rr = n, ok = -1;
    while (ll <= rr) {
      int mid = (ll + rr) >> 1;
      if (sol(mid) >= i) {
        ok = mid;
        ll = mid + 1;
      } else {
        rr = mid - 1;
      }
    }
    while (pos <= ok) {
      ans[pos] = i;
      pos++;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
