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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 300005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
vector<int> V[N];
int n, dp[N][21], h[N][21], tmp[N], len[N], tot, x, y;
long long ans = 0;
inline bool cmp(int x, int y) { return x > y; }
void dfs(int x, int f) {
  for (__typeof((V[x]).begin()) it = (V[x]).begin(); it != (V[x]).end(); it++)
    if (*it != f)
      dfs(*it, x), len[x] = ((len[x]) > (len[*it]) ? (len[x]) : (len[*it]));
  ans += ++len[x];
  dp[x][1] = h[x][1] = n;
  for (int p = (2); p <= (20); ++p) {
    tot = 0;
    for (__typeof((V[x]).begin()) it = (V[x]).begin(); it != (V[x]).end(); it++)
      if (*it != f) tmp[++tot] = dp[*it][p - 1];
    sort(tmp + 1, tmp + 1 + tot, cmp);
    for (int k = (tot); k >= (2); --k)
      if (tmp[k] >= k) {
        h[x][p] = dp[x][p] = k;
        break;
      }
    for (__typeof((V[x]).begin()) it = (V[x]).begin(); it != (V[x]).end(); it++)
      if (*it != f)
        h[x][p] = ((h[x][p]) > (h[*it][p]) ? (h[x][p]) : (h[*it][p]));
  }
  for (int i = (1); i <= (20); ++i) {
    ans += (long long)i * (h[x][i] - h[x][i + 1]);
    if (!h[x][i + 1]) {
      ans -= i;
      break;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i)
    scanf("%d%d", &x, &y), V[x].push_back(y), V[y].push_back(x);
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}
