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 N = 3e5;
int n, dp[N][19];
long long ans;
vector<int> g[N];
int intial(int u, int p) {
  int ret = 0;
  for (auto i : g[u])
    if (i != p) ret = max(ret, intial(i, u));
  ans += ret + 1;
  return ret + 1;
}
void dfs(int u, int p) {
  bool leaf = 1;
  dp[u][1] = n;
  for (auto i : g[u])
    if (i != p) {
      leaf = 0;
      dfs(i, u);
    }
  if (leaf) return;
  for (int i = 2; i < 19; i++) {
    vector<int> v;
    for (auto j : g[u])
      if (j != p && dp[j][i - 1] != -1) v.push_back(dp[j][i - 1]);
    sort(v.rbegin(), v.rend());
    for (int j = 0; j < v.size(); j++) {
      if (v[j] <= j) break;
      dp[u][i] = j + 1;
    }
  }
}
void find(int u, int p) {
  for (auto i : g[u])
    if (i != p) {
      find(i, u);
      for (int j = 0; j < 19; j++) dp[u][j] = max(dp[u][j], dp[i][j]);
    }
  int did = 1;
  for (int i = 18; i; i--)
    if (dp[u][i] != -1 && dp[u][i] > did) {
      ans += i * 1LL * (dp[u][i] - did);
      did = dp[u][i];
    }
}
int main() {
  scanf("%d", &n);
  memset(dp, -1, sizeof dp);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    g[--a].push_back(--b);
    g[b].push_back(a);
  }
  intial(0, -1);
  dfs(0, -1);
  find(0, -1);
  printf("%lld\n", ans);
}
