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;
struct edgeType {
  int to, nxt;
};
edgeType e[100005 << 1];
int head[100005];
int n;
int dfn[100005], parent[100005];
int rk[100005];
int dp[100005], qhy[100005];
int ans[100005];
void add_e(int u, int v, int x) {
  e[x].to = v;
  e[x].nxt = head[u];
  head[u] = x;
}
void dfs(int x, int p) {
  int i;
  parent[x] = p;
  dfn[x] = ++dfn[0];
  rk[x] = x;
  ans[x] = -1;
  for (i = head[x]; i; i = e[i].nxt)
    if (e[i].to != p) dfs(e[i].to, x);
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
void standing_by() {
  int i, u, v;
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_e(u, v, i * 2 - 1);
    add_e(v, u, i * 2);
  }
  dfs(1, 0);
  sort(rk + 1, rk + 1 + n, cmp);
}
void get_ans(int i) {
  if (ans[i] >= 0) return;
  int j, k;
  int x, y, z;
  for (j = n; j >= 1; j--) {
    x = rk[j];
    y = z = 0;
    for (k = head[x]; k; k = e[k].nxt) {
      if (e[k].to == parent[x]) continue;
      dp[x] += dp[e[k].to];
      if (qhy[e[k].to] >= y) {
        z = y;
        y = qhy[e[k].to];
      } else if (qhy[e[k].to] >= z) {
        z = qhy[e[k].to];
      }
      dp[e[k].to] = 0;
      qhy[e[k].to] = 0;
    }
    if (y + z + 1 >= i) {
      dp[x]++;
    } else {
      qhy[x] = y + 1;
    }
  }
  ans[i] = dp[1];
  dp[1] = qhy[1] = 0;
}
void complete() {
  int i;
  int l, r, z;
  int boundary = sqrt(n);
  for (i = 1; i <= boundary; i++) get_ans(i);
  for (i = boundary + 1; i <= n;) {
    get_ans(i);
    l = i;
    r = n + 1;
    while (r - l > 1) {
      z = (l + r) >> 1;
      get_ans(z);
      if (ans[z] == ans[i])
        l = z;
      else
        r = z;
    }
    for (; i <= l; i++) ans[i] = ans[l];
  }
  for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
}
int main() {
  standing_by();
  complete();
  return 0;
}
