You're given a tree with n vertices rooted at 1.

We say that there's a k-ary heap of depth m located at u if the following holds:

  * For m = 1 u itself is a k-ary heap of depth 1. 
  * For m > 1 vertex u is a k-ary heap of depth m if at least k of its children are k-ary heaps of depth at least m - 1. 



Denote dpk(u) as maximum depth of k-ary heap in the subtree of u (including u). Your goal is to compute <image>.

Input

The first line contains an integer n denoting the size of the tree (2 ≤ n ≤ 3·105). 

The next n - 1 lines contain two integers u, v each, describing vertices connected by i-th edge.

It's guaranteed that the given configuration forms a tree.

Output

Output the answer to the task.

Examples

Input

4
1 3
2 3
4 3


Output

21


Input

4
1 2
2 3
3 4


Output

22

Note

Consider sample case one.

For k ≥ 3 all dpk will be equal to 1.

For k = 2 dpk is 2 if <image> and 1 otherwise.

For k = 1 dpk values are (3, 1, 2, 1) respectively.

To sum up, 4·1 + 4·1 + 2·2 + 2·1 + 3 + 1 + 2 + 1 = 21.
#include <bits/stdc++.h>
using std::vector;
const int N = 300005;
int n, f[N][30], g[N][30], tmp[N];
long long ans;
vector<int> e[N];
int dfs(int u, int fa) {
  int sz = 0, mx = 0;
  for (int v : e[u])
    if (v != fa) mx = std::max(mx, dfs(v, u)), ++sz;
  ++mx;
  f[u][1] = n;
  ans += n - 1 + mx;
  for (int i = 2; i <= 25; ++i) {
    std::fill(tmp, tmp + sz + 1, 0);
    for (int v : e[u])
      if (v != fa) ++tmp[std::min(f[v][i - 1], sz)];
    for (int j = sz, cur = 0; ~j; --j) {
      cur += tmp[j];
      if (cur >= j) {
        f[u][i] = j;
        break;
      }
    }
    g[u][i] = f[u][i];
    for (int v : e[u])
      if (v != fa) g[u][i] = std::max(g[u][i], g[v][i]);
    ans += std::max(g[u][i] - 1, 0);
  }
  return mx;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i)
    scanf("%d%d", &x, &y), e[x].push_back(y), e[y].push_back(x);
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}
