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;
vector<int> g[300005];
int dp[300005][25];
int D[300005];
int n;
void dfs1(int u, int p) {
  D[u] = 1;
  for (int i = 0; i < (int)(g[u]).size(); i++) {
    int v = g[u][i];
    if (v != p) {
      dfs1(v, u);
      D[u] = max(D[u], D[v] + 1);
    }
  }
  for (int d = 2; d < 20; d++) {
    vector<int> sum;
    for (int i = 0; i < (int)(g[u]).size(); i++) {
      int v = g[u][i];
      if (v != p) sum.emplace_back(dp[v][d - 1]);
    }
    sort((sum).begin(), (sum).end());
    reverse((sum).begin(), (sum).end());
    int mn = 1e9;
    for (int i = 0; i < (int)(sum).size(); i++) {
      mn = min(mn, sum[i]);
      dp[u][d] = max(dp[u][d], min(i + 1, mn));
    }
  }
  dp[u][1] = n;
}
void dfs2(int u, int p) {
  for (int i = 0; i < (int)(g[u]).size(); i++) {
    int v = g[u][i];
    if (v != p) {
      dfs2(v, u);
      for (int d = 1; d < 20; d++) {
        dp[u][d] = max(dp[u][d], dp[v][d]);
      }
    }
  }
  for (int d = 18; d >= 1; d--) dp[u][d] = max(dp[u][d], dp[u][d + 1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].emplace_back(b);
    g[b].emplace_back(a);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int u = 1; u <= n; u++) {
    if (D[u] > 19) {
      ans += D[u];
    }
    for (int d = 19; d >= 1; d--) {
      int diff = dp[u][d] - dp[u][d + 1];
      if (D[u] > 19 && d == 19) diff--;
      ans += diff * 1ll * d;
    }
  }
  printf("%lld\n", ans);
}
