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;
long long ans;
int ne[N << 1], fi[N], zz[N << 1], tot, x, y, dp[N][25], T, n, Max[N],
    dp2[N][25];
void jb(int x, int y) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
}
int cmp(int x, int y) { return dp[x][T] > dp[y][T]; }
void dfs(int x, int y) {
  Max[x] = 1;
  vector<int> v;
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != y) {
      dfs(zz[i], x);
      Max[x] = max(Max[x], Max[zz[i]] + 1);
      v.push_back(zz[i]);
    }
  dp[x][0] = 1e9;
  ans += n - 1;
  for (int i = 1; i <= 20; i++) {
    T = i - 1;
    sort(v.begin(), v.end(), cmp);
    for (int j = 0; j < (int)v.size(); j++)
      if (dp[v[j]][i - 1] >= j + 1) dp[x][i] = j + 1;
    dp2[x][i] = dp[x][i];
    for (int j : v) dp2[x][i] = max(dp2[x][i], dp2[j][i]);
    if (dp2[x][i] > 1) ans += dp2[x][i] - 1;
  }
  ans += Max[x];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    jb(x, y);
    jb(y, x);
  }
  dfs(1, 0);
  printf("%lld\n", ans);
  return 0;
}
