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;
template <typename T>
bool chkmax(T &a, T b) {
  return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T &a, T b) {
  return (a > b) ? a = b, 1 : 0;
}
inline int read() {
  int x = 0, fh = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fh = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return x * fh;
}
int n;
int he[400005], to[1200005], ne[1200005], e;
void add(int x, int y) {
  to[++e] = y;
  ne[e] = he[x];
  he[x] = e;
}
long long dp[400005][20];
long long dep[400005];
long long num[400005], tmp;
long long ans;
void dfs(int x, int ff) {
  dep[x] = 1;
  for (int i = he[x]; i; i = ne[i]) {
    int v = to[i];
    if (v == ff) continue;
    dfs(v, x);
    chkmax(dep[x], dep[v] + 1);
  }
  dp[x][1] = n;
  for (int i = (2), _end_ = (int)(19); i <= _end_; ++i) {
    tmp = 0;
    for (int j = he[x]; j; j = ne[j]) {
      int v = to[j];
      if (v == ff) continue;
      if (dp[v][i - 1]) {
        num[++tmp] = dp[v][i - 1];
      }
    }
    sort(num + 1, num + tmp + 1);
    for (int j = (tmp), _end_ = (int)(1); j >= _end_; --j) {
      if (num[tmp - j + 1] >= j) {
        dp[x][i] = j;
        break;
      }
    }
  }
}
void dfs2(int x, int ff) {
  for (int i = he[x]; i; i = ne[i]) {
    int v = to[i];
    if (v == ff) continue;
    dfs2(v, x);
    for (int j = (2), _end_ = (int)(19); j <= _end_; ++j)
      chkmax(dp[x][j], dp[v][j]);
  }
}
int main() {
  n = read();
  for (int i = (1), _end_ = (int)(n - 1); i <= _end_; ++i) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) ans += dep[i];
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i)
    for (int j = (1), _end_ = (int)(19); j <= _end_; ++j) {
      if (dp[i][j]) ans += dp[i][j] - 1;
    }
  printf("%lld\n", ans);
  return 0;
}
