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, head[300005], num, x, y, dep[300005], Max[300005][21], own[300005][21];
long long ans;
struct edge {
  int to, nxt;
} a[300005 << 1];
void add(int x, int y) { a[++num] = (edge){y, head[x]}, head[x] = num; }
bool check(int x, int f, int id, int y) {
  int res = 0;
  for (int e = head[x]; e; e = a[e].nxt) {
    int k = a[e].to;
    if (k == f) continue;
    res += (own[k][id - 1] >= y);
  }
  return res >= y;
}
void dfs(int x, int f) {
  dep[x] = 1;
  own[x][1] = Max[x][1] = n;
  for (int e = head[x]; e; e = a[e].nxt) {
    int k = a[e].to;
    if (k == f) continue;
    dfs(k, x);
    dep[x] = max(dep[x], dep[k] + 1);
    for (int j = 2; j <= 20; ++j) {
      Max[x][j] = max(Max[x][j], Max[k][j]);
    }
  }
  int last = n;
  for (int j = 2; j <= 20; ++j) {
    int l = 1, r = last, mid;
    while (l < r) {
      mid = (l + r + 1) >> 1;
      if (check(x, f, j, mid))
        l = mid;
      else
        r = mid - 1;
    }
    own[x][j] = l;
    Max[x][j] = max(Max[x][j], l);
    last = l;
  }
  ans += dep[x];
  for (int j = 1; j < 20; ++j) {
    ans += j * (Max[x][j] - Max[x][j + 1]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}
