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 LIMIT = 20;
int n, dep[300001];
vector<int> adj[300001];
int dp[300001][LIMIT], dp2[300001][LIMIT];
void dfs(int x, int pre) {
  vector<int> child;
  for (auto y : adj[x]) {
    if (y != pre) {
      dfs(y, x);
      child.push_back(y);
      dep[x] = max(dep[x], dep[y] + 1);
    }
  }
  dp[x][1] = dp2[x][1] = n;
  for (int i = 2; i < LIMIT; ++i) {
    int lo = 0, hi = n;
    while (lo < hi) {
      int md = lo + hi + 1 >> 1, tot = 0;
      for (auto y : child) {
        if (dp[y][i - 1] >= md) {
          ++tot;
        }
      }
      (tot >= md) ? lo = md : hi = md - 1;
    }
    dp[x][i] = dp2[x][i] = lo + hi >> 1;
    for (auto y : child) {
      dp2[x][i] = max(dp2[x][i], dp2[y][i]);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y), --x, --y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, -1);
  long long ans = 0LL;
  for (int i = 0; i < n; ++i) {
    ans += dep[i];
  }
  for (int u = 0; u < n; ++u) {
    for (int i = 1; i < LIMIT; ++i) {
      int lo = 2, hi = dp2[u][i];
      if (i + 1 < LIMIT) {
        lo = max(lo, dp2[u][i + 1] + 1);
      }
      if (lo <= hi) {
        ans += (long long)(hi - lo + 1) * (i - 1);
      }
    }
  }
  ans += (long long)n * n;
  cout << ans << endl;
}
