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 namespace std;
const int N = 300005;
vector<int> e[N];
int n, v, f[N], g[N], q[N];
long long sum;
void dfs(int x, int fa, int v) {
  f[x] = 1;
  g[x] = 1;
  for (auto i : e[x])
    if (i != fa) {
      dfs(i, x, v);
      g[x] = max(g[x], g[i]);
    }
  if (e[x].size() - (fa != 0) >= v) {
    *q = 0;
    for (auto i : e[x])
      if (i != fa) q[++*q] = -f[i];
    assert(*q >= v);
    nth_element(q + 1, q + v, q + *q + 1);
    f[x] = -q[v];
    g[x] = max(g[x], ++f[x]);
  }
  sum += g[x];
}
void dfs2(int x, int fa, int v) {
  f[x] = e[x].size() - (fa != 0);
  g[x] = 0;
  for (auto i : e[x])
    if (i != fa) {
      dfs2(i, x, v);
      f[x] = max(f[x], f[i]);
      g[x] = max(g[x], g[i]);
    }
  *q = 0;
  for (auto i : e[x])
    if (i != fa) q[++*q] = e[i].size() - 1;
  sort(q + 1, q + *q + 1);
  reverse(q + 1, q + *q + 1);
  int tmp = 0;
  for (; tmp < *q && q[tmp + 1] >= tmp + 1; ++tmp)
    ;
  g[x] = max(g[x], tmp);
  sum += n - v + 1;
  if (f[x] >= v) sum += f[x] - v + 1;
  if (g[x] >= v) sum += g[x] - v + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  for (v = 1; v * v * v <= n; v++) dfs(1, 0, v);
  dfs2(1, 0, v);
  printf("%lld\n", sum);
}
