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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
vector<int> mp[100005], mp1[100005];
int dp[100005], g[100005], cnt, ans[100005];
void bd(int x, int pa) {
  g[x] = ++cnt;
  for (int c : mp[x])
    if (c != pa) {
      bd(c, x);
      mp1[g[x]].push_back(g[c]);
    }
}
void dfs(int k) {
  for (int x = n; x; x--) {
    int m1 = 0, m2 = 0;
    dp[x] = g[x] = 0;
    for (int c : mp1[x]) {
      dp[x] += dp[c];
      if (g[c] > m1) {
        m2 = m1;
        m1 = g[c];
      } else if (g[c] > m2)
        m2 = g[c];
    }
    if (m1 + m2 + 1 >= k) {
      dp[x]++;
    } else
      g[x] = m1 + 1;
  }
}
void cal(int l, int r, int L, int R) {
  if (r < l) return;
  if (L == R) {
    for (int i = l; i <= r; i++) ans[i] = L;
    return;
  }
  int mid = (l + r) / 2;
  dfs(mid);
  ans[mid] = dp[1];
  cal(l, mid - 1, ans[mid], R);
  cal(mid + 1, r, L, ans[mid]);
}
void fmain(int ID) {
  scanf("%d", &n);
  for (int i = 0, u, v; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    mp[u].push_back(v);
    mp[v].push_back(u);
  };
  bd(1, 0);
  cal(1, n, 0, n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) printf("%d\n", ans[i]);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
