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 maxn = 3e5;
int dp[maxn + 5][25];
vector<int> g[maxn + 5];
int a[maxn + 5], mx[maxn + 5];
int len;
int n;
long long ans = 0;
void dfs(int k, int fa) {
  for (auto u : g[k]) {
    if (u == fa) continue;
    dfs(u, k);
  }
  dp[k][1] = n;
  for (int dep = 2; dep <= 20; ++dep) {
    len = 0;
    for (auto u : g[k]) {
      if (u == fa) continue;
      a[++len] = dp[u][dep - 1];
    }
    sort(a + 1, a + len + 1);
    reverse(a + 1, a + len + 1);
    for (int i = len; i >= 1; --i)
      if (a[i] >= i) {
        dp[k][dep] = i;
        break;
      }
  }
}
void dfs1(int k, int fa) {
  mx[k] = 1;
  for (auto u : g[k]) {
    if (u == fa) continue;
    dfs1(u, k);
    for (int i = 1; i <= 20; ++i) dp[k][i] = max(dp[k][i], dp[u][i]);
    mx[k] = max(mx[k], mx[u] + 1);
  }
}
int main() {
  scanf("%d", &n);
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y), g[y].push_back(x);
  }
  dfs(1, 0);
  dfs1(1, 0);
  for (int i = 1; i <= n; ++i) ans += mx[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = 20; j >= 1; --j) dp[i][j] = max(dp[i][j], dp[i][j + 1]);
    for (int j = 1; j <= 21; ++j)
      if (dp[i][j] == 0) {
        dp[i][j] = 1;
        break;
      }
    for (int j = 1; j <= 20; ++j) {
      if (dp[i][j] == 1) break;
      ans += 1LL * j * (dp[i][j] - dp[i][j + 1]);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
