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;
int N;
vector<int> graph[300005];
int mdp[20][300005];
int cdp[20][300005];
int mdep[300005];
int fre[300005];
long long ans = 0;
void dfs(int n) {
  cdp[1][n] = N;
  mdep[n] = 1;
  for (int e : graph[n]) {
    graph[e].erase(find(graph[e].begin(), graph[e].end(), n));
    dfs(e);
    mdep[n] = max(mdep[n], mdep[e] + 1);
    for (int k = 1; k < 20; k++) {
      mdp[k][n] = max(mdp[k][n], mdp[k][e]);
    }
  }
  int dgr = graph[n].size();
  for (int k = 2; k < 20; k++) {
    fill(fre, fre + dgr + 2, 0);
    for (int e : graph[n]) {
      fre[min(dgr, cdp[k - 1][e])]++;
    }
    for (int d = dgr; d; d--) {
      fre[d] += fre[d + 1];
      if (fre[d] >= d) {
        cdp[k][n] = d;
        break;
      }
    }
  }
  ans += mdep[n];
  for (int k = 1; k < 20; k++) {
    mdp[k][n] = max(mdp[k][n], cdp[k][n]);
    if (mdp[k][n] >= 2) {
      ans += mdp[k][n] - 1;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  for (int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  dfs(1);
  cout << ans;
}
