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;
long long int Ans = 0;
int Par[300005];
int Reach[300005];
int dp[300005][25];
vector<int> adj[300005];
vector<pair<int, int> > Update[300005];
long long int dfs1(int u, int p) {
  long long int mx = 0;
  Par[u] = p;
  for (int i = 0; i < adj[u].size(); i++) {
    int w = adj[u][i];
    if (w == p) continue;
    long long int ret = dfs1(w, u);
    mx = max(mx, ret);
  }
  Ans += mx + 1;
  return mx + 1;
}
void solve(int u, int p) {
  for (int i = 0; i < adj[u].size(); i++) {
    int w = adj[u][i];
    if (w == p) continue;
    solve(w, u);
  }
  dp[u][1] = n;
  for (int d = 2; d < 25; d++) {
    vector<int> v;
    for (int i = 0; i < adj[u].size(); i++) {
      int w = adj[u][i];
      if (w == p) continue;
      if (dp[w][d - 1]) v.push_back(dp[w][d - 1]);
    }
    sort(v.begin(), v.end(), greater<int>());
    int mx = 0;
    for (int i = 1; i <= v.size(); i++)
      if (v[i - 1] >= i) mx = i;
    dp[u][d] = mx;
    if (mx) Update[mx].push_back(make_pair(u, d));
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs1(1, 0);
  solve(1, 0);
  long long int Add = n;
  for (int i = 1; i <= n; i++) Reach[i] = 1;
  for (int k = n; k > 1; k--) {
    for (int i = 0; i < Update[k].size(); i++) {
      int u = Update[k][i].first;
      int d = Update[k][i].second;
      while (u && Reach[u] < d) {
        Add += d - Reach[u];
        Reach[u] = d;
        u = Par[u];
      }
    }
    Ans += Add;
  }
  cout << Ans << endl;
}
