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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e9 + 7;
const long long max_n = 3e5 + 3;
long long n;
vector<long long> gr[max_n];
vector<long long> dp[max_n];
long long mx_k[max_n];
void dfs(long long v, long long pr) {
  vector<long long> ch;
  for (long long to : gr[v]) {
    if (to == pr) continue;
    dfs(to, v);
    ch.emplace_back(to);
  }
  vector<vector<long long>> a((long long)(ch.size()));
  for (long long to : ch) {
    for (long long i = 0; i < min((long long)(ch.size()), mx_k[to]); i++) {
      if ((long long)(dp[to].size()) > i)
        a[i].emplace_back(dp[to][i]);
      else
        a[i].emplace_back(2);
    }
  }
  for (long long k = 1; k <= (long long)(ch.size()); k++) {
    sort(a[k - 1].begin(), a[k - 1].end());
    reverse(a[k - 1].begin(), a[k - 1].end());
    if ((long long)(a[k - 1].size()) > k - 1)
      dp[v].emplace_back(a[k - 1][k - 1] + 1);
  }
  mx_k[v] = (long long)(ch.size());
}
void dfs1(long long v, long long pr) {
  for (long long to : gr[v]) {
    if (to == pr) continue;
    dfs1(to, v);
    for (long long k = 0; k < (long long)(dp[to].size()); k++) {
      if ((long long)(dp[v].size()) == k)
        dp[v].emplace_back(dp[to][k]);
      else
        dp[v][k] = max(dp[v][k], dp[to][k]);
    }
    mx_k[v] = max(mx_k[v], mx_k[to]);
  }
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long v1, v2;
    cin >> v1 >> v2;
    gr[--v1].emplace_back(--v2);
    gr[v2].emplace_back(v1);
  }
  dfs(0, -1);
  dfs1(0, -1);
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j : dp[i]) res += j;
    res += (mx_k[i] - (long long)(dp[i].size())) * 2 + (n - mx_k[i]);
  }
  cout << res;
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cout.precision(10);
  solve();
}
