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 LG = 20;
int dp[300100][LG], mx[300100][LG], dep[300100];
int buf[300100];
vector<int> g[300100];
long long solve(int u, int p) {
  long long ans = 0;
  dep[u] = 1;
  for (auto v : g[u]) {
    if (v == p) continue;
    ans += solve(v, u);
    dep[u] = max(dep[u], dep[v] + 1);
    for (int j = 0; j < LG; j++) {
      mx[u][j] = max(mx[u][j], mx[v][j]);
    }
  }
  for (int j = 0; j < LG; j++) {
    if (j) {
      int m = 0;
      for (auto v : g[u]) {
        if (v != p) {
          buf[m++] = dp[v][j - 1];
        }
      }
      sort(buf, buf + m, greater<int>());
      while (m && buf[m - 1] < m) m--;
      dp[u][j] = m;
    }
    mx[u][j] = max(mx[u][j], dp[u][j]);
    ans += mx[u][j];
  }
  if (dep[u] > LG) ans += dep[u] - LG;
  return ans;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) dp[i][0] = n;
  printf("%lld\n", solve(1, 0));
  return 0;
}
