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;
vector<int> g[300005];
int n;
int f[300005][32];
int w[300005];
int a[300005];
int m;
long long r;
void dfs(int x, int p) {
  w[x] = 1;
  for (int y : g[x])
    if (y != p) {
      dfs(y, x);
      w[x] = max(w[x], 1 + w[y]);
    }
  f[x][1] = n;
  for (int d = 2, ThxDem = 21; d < ThxDem; ++d) {
    m = 0;
    for (int y : g[x])
      if (y != p) a[m++] = f[y][d - 1];
    sort(a, a + m);
    reverse(a, a + m);
    f[x][d] = 0;
    while (f[x][d] < m && a[f[x][d]] > f[x][d]) f[x][d]++;
    f[x][d] = max(f[x][d], 1);
  }
  f[x][21] = 1;
}
void dfs2(int x, int p) {
  for (int y : g[x])
    if (y != p) {
      dfs2(y, x);
      for (int d = 1, ThxDem = 21; d < ThxDem; ++d)
        f[x][d] = max(f[x][d], f[y][d]);
    }
}
int main() {
  scanf("%d", &n);
  for (int _ = 1, ThxDem = n; _ < ThxDem; ++_) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0, -1);
  dfs2(0, -1);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) {
    r += w[i];
    for (int d = 1, ThxDem = 21; d < ThxDem; ++d)
      r += d * (f[i][d] - f[i][d + 1]);
  }
  printf("%lld\n", r);
  return 0;
}
