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[300010], ch[300010];
vector<pair<int, int> > up[300010];
int p[300010], dp[300010][30], val[300010];
int u, v, n;
long long ans;
void dfs1(int u, int f) {
  p[u] = f;
  for (int v : g[u])
    if (v != f) {
      ch[u].emplace_back(v);
      dfs1(v, u);
    }
}
void dfs2(int u) {
  for (int v : ch[u]) dfs2(v);
  auto &_dp = dp[u];
  _dp[1] = n;
  for (int k = 2; k < 30; k++) {
    vector<int> hp;
    for (int v : ch[u])
      if (dp[v][k - 1]) hp.emplace_back(dp[v][k - 1]);
    sort(hp.begin(), hp.end(), greater<int>());
    for (int s = hp.size() - 1; s > 0; s--)
      if (hp[s] >= s + 1) {
        _dp[k] = s + 1;
        break;
      }
    if (_dp[k]) up[_dp[k]].emplace_back(u, k);
  }
}
int dfs3(int u) {
  if (!ch[u].size()) {
    ans++;
    return 1;
  }
  int mx = 0;
  for (int v : ch[u]) mx = max(mx, dfs3(v));
  ans += mx + 1;
  return mx + 1;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs1(1, 0);
  fill_n(&dp[0][0], 300010 * 30, 0);
  dfs2(1);
  dfs3(1);
  long long add = n;
  fill_n(val, 300010, 1);
  for (int k = n; k > 1; k--) {
    for (auto e : up[k]) {
      int u = e.first, v = e.second;
      while (u && val[u] < v) {
        add += v - val[u];
        val[u] = v;
        u = p[u];
      }
    }
    ans += add;
  }
  cout << ans << endl;
  return 0;
}
