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 maxn = 3e5 + 5;
const int lg = 21;
int n, dp[maxn][lg], k, mx[maxn], d[lg];
long long s;
vector<int> ad[maxn];
void fix(int u, int p) {
  for (auto it = ad[u].begin(); it != ad[u].end(); ++it) {
    if (*it == p) {
      ad[u].erase(it);
      break;
    }
  }
  for (auto v : ad[u]) {
    fix(v, u);
    mx[u] = max(mx[u], mx[v]);
  }
  mx[u]++;
  s += mx[u];
}
void dfs(int u) {
  for (auto v : ad[u]) dfs(v);
  dp[u][1] = n;
  for (int h = 2; h < lg + 1; ++h) {
    vector<int> tmp;
    for (auto v : ad[u]) {
      if (dp[v][h - 1]) {
        tmp.push_back(dp[v][h - 1]);
      }
    }
    sort(tmp.begin(), tmp.end());
    for (int i = 0; i < tmp.size(); ++i) {
      if (tmp[i] >= tmp.size() - i) {
        dp[u][h] = tmp.size() - i;
        break;
      }
    }
  }
}
void dsf(int u) {
  for (auto v : ad[u]) {
    dsf(v);
  }
  for (int h = 2; h < lg + 1; ++h) {
    for (auto v : ad[u]) {
      if (dp[v][h]) dp[u][h] = max(dp[u][h], dp[v][h]);
    }
    d[h] = dp[u][h];
  }
  d[1] = n;
  int t = 0;
  for (int h = lg; h >= 1; h--)
    if (d[h]) {
      d[h] -= t;
      s += h * d[h];
      if (!t) s -= h;
      t += d[h];
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    ad[x].push_back(y);
    ad[y].push_back(x);
  }
  fix(1, -1);
  dfs(1);
  dsf(1);
  cout << s;
  return 0;
}
