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;
vector<int> G[300005];
int PosRes[22][300005], Pos[22][300005];
int V[22][300005];
int DP[300005];
vector<int> Who;
long long ans;
int Cnt[30];
void Read() {
  scanf("%d", &N);
  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);
  }
}
void DFS(int node, int father) {
  Pos[1][node] = PosRes[1][node] = N;
  for (int j = 2; (1 << (j - 1)) <= N; j++) Pos[j][node] = PosRes[j][node] = 1;
  DP[node] = 1;
  for (int i = 0; i < G[node].size(); i++) {
    int neighb = G[node][i];
    if (neighb == father) continue;
    DFS(neighb, node);
    DP[node] = max(DP[node], DP[neighb] + 1);
  }
  for (int j = 1; (1 << (j - 1)) <= N; j++) Cnt[j] = 0;
  for (int i = 0; i < G[node].size(); i++) {
    int neighb = G[node][i];
    if (neighb == father) continue;
    for (int j = 1; (1 << (j - 1)) <= N; j++) {
      PosRes[j][node] = max(PosRes[j][node], PosRes[j][neighb]);
      V[j][++Cnt[j]] = Pos[j][neighb];
    }
  }
  for (int j = 1; (1 << (j - 1)) <= N; j++) {
    sort(V[j] + 1, V[j] + Cnt[j] + 1);
  }
  for (int j = 2; (1 << (j - 1)) <= N; j++) {
    int ind = Cnt[j - 1] - Pos[j][node] + 1;
    while (Pos[j][node] <= N) {
      ind--;
      if (ind <= 0) break;
      if (Pos[j][node] + 1 <= V[j - 1][ind]) {
        ++Pos[j][node];
      } else
        break;
    }
    PosRes[j][node] = max(PosRes[j][node], Pos[j][node]);
  }
  for (int j = 1; (1 << (j - 1)) <= N; j++) {
    int next = (1 << j) > N ? 1 : PosRes[j + 1][node];
    ans += 1LL * (PosRes[j][node] - next) * j;
  }
}
int main() {
  Read();
  DFS(1, 0);
  for (int i = 1; i <= N; i++) ans += DP[i];
  printf("%I64d\n", ans);
  return 0;
}
