#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 1e6 + 1000;
int n, tot, ans, len;
int head[N], Ans[N], f[N], g[N];
struct node {
  int nex, to;
} e[N << 1];
void add(int f, int t) {
  e[++tot].nex = head[f];
  e[tot].to = t;
  head[f] = tot;
}
void dfs(int now, int fa) {
  f[now] = g[now] = 1;
  for (int i = head[now], to; i; i = e[i].nex) {
    to = e[i].to;
    if (to == fa) continue;
    dfs(to, now);
    if (f[to] + 1 > f[now])
      g[now] = f[now], f[now] = f[to] + 1;
    else if (f[to] + 1 > g[now])
      g[now] = f[to] + 1;
  }
  if (f[now] + g[now] - 1 >= len) ans++, f[now] = 0;
}
void solve(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;
  ans = 0;
  len = mid;
  dfs(1, 0);
  Ans[mid] = ans;
  solve(l, mid - 1, ans, R);
  solve(mid + 1, r, L, ans);
}
int main() {
  n = read();
  for (int i = 1, x, y; i < n; i++) {
    x = read();
    y = read();
    add(x, y);
    add(y, x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) cout << Ans[i] << "\n";
  fclose(stdin);
  fclose(stdout);
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 1e6 + 1000;
int n, tot, ans, len;
int head[N], Ans[N], f[N], g[N];
struct node {
  int nex, to;
} e[N << 1];
void add(int f, int t) {
  e[++tot].nex = head[f];
  e[tot].to = t;
  head[f] = tot;
}
void dfs(int now, int fa) {
  f[now] = g[now] = 1;
  for (int i = head[now], to; i; i = e[i].nex) {
    to = e[i].to;
    if (to == fa) continue;
    dfs(to, now);
    if (f[to] + 1 > f[now])
      g[now] = f[now], f[now] = f[to] + 1;
    else if (f[to] + 1 > g[now])
      g[now] = f[to] + 1;
  }
  if (f[now] + g[now] - 1 >= len) ans++, f[now] = 0;
}
void solve(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;
  ans = 0;
  len = mid;
  dfs(1, 0);
  Ans[mid] = ans;
  solve(l, mid - 1, ans, R);
  solve(mid + 1, r, L, ans);
}
int main() {
  n = read();
  for (int i = 1, x, y; i < n; i++) {
    x = read();
    y = read();
    add(x, y);
    add(y, x);
  }
  solve(1, n, 0, n);
  for (int i = 1; i <= n; i++) cout << Ans[i] << "\n";
  fclose(stdin);
  fclose(stdout);
  return 0;
}

```