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>
const int MAXN = 2e6 + 7;
int n, m, q;
int home[MAXN], next[MAXN], to[MAXN];
int ans[MAXN];
int dp[MAXN], ccnt, tmp, cnt;
inline void ct(const int& x, const int& y) {
  ccnt++;
  next[ccnt] = home[x];
  home[x] = ccnt;
  to[ccnt] = y;
  return;
}
inline void dfs(int u, int fa, int len) {
  dp[u] = 0;
  int maxx1 = 0;
  int maxx2 = 0;
  for (int i = home[u]; i; i = next[i]) {
    int v = to[i];
    if (v == fa) {
      continue;
    }
    dfs(v, u, len);
    if (dp[v] > maxx1) {
      maxx2 = maxx1;
      maxx1 = dp[v];
    } else if (dp[v] > maxx2) {
      maxx2 = dp[v];
    }
  }
  if (maxx1 + maxx2 + 1 >= len) {
    cnt++;
    dp[u] = 0;
  } else {
    dp[u] = maxx1 + 1;
  }
  return;
}
inline void Dic(int l, int r, int L, int R) {
  if (l > r || L > R) return;
  if (L == R) {
    for (int i = l; i <= r; i++) {
      ans[i] = L;
    }
    return;
  }
  int mid = (l + r) >> 1;
  cnt = 0;
  dfs(1, 0, mid);
  ans[mid] = cnt;
  Dic(l, mid - 1, cnt, R);
  Dic(mid + 1, r, L, cnt);
  return;
}
int main() {
  scanf("%d", &n);
  int x, y;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    ct(x, y);
    ct(y, x);
  }
  Dic(1, n, 0, n);
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
