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 N = 3e5 + 5;
const int mod = 1e9 + 7;
long long ans;
int n, d[N], son[N], size[N];
vector<int> G[N], dp[N];
bool cmp(int i, int j) { return d[i] > d[j]; }
void dfs(int x, int f) {
  size[x] = 1;
  if (find(G[x].begin(), G[x].end(), f) != G[x].end())
    G[x].erase(find(G[x].begin(), G[x].end(), f));
  d[x] = G[x].size();
  dp[x].resize(d[x] + 1);
  for (int u : G[x]) {
    dfs(u, x);
    size[x] += size[u];
    if (size[u] >= size[son[x]]) son[x] = u;
  }
  sort(G[x].begin(), G[x].end(), cmp);
  for (int i = 1; i <= d[x]; ++i) {
    multiset<int> S;
    dp[x][i] = 2;
    for (int u : G[x]) {
      if (d[u] < i) break;
      S.insert(dp[u][i]);
    }
    while (S.size() > i) S.erase(S.begin());
    if (S.size() == i) dp[x][i] = (*S.begin()) + 1;
  }
}
int f[N];
long long sum;
void clear(int x) {
  for (int i = 1; i <= d[x]; ++i) f[i] = 1;
  for (int u : G[x]) clear(u);
}
void insert(int x) {
  for (int i = 1; i <= d[x]; ++i) {
    if (dp[x][i] > f[i]) {
      sum += dp[x][i] - f[i];
      f[i] = dp[x][i];
    }
  }
  for (int u : G[x]) insert(u);
}
void dfs(int x) {
  for (int u : G[x]) {
    if (u == son[x]) continue;
    dfs(u);
    sum = n;
    clear(u);
  }
  if (son[x]) dfs(son[x]);
  for (int u : G[x])
    if (u != son[x]) insert(u);
  for (int i = 1; i <= d[x]; ++i) {
    if (dp[x][i] > f[i]) {
      sum += dp[x][i] - f[i];
      f[i] = dp[x][i];
    }
  }
  ans += sum;
}
int main() {
  cin >> 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);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) f[i] = 1;
  sum = n;
  dfs(1);
  cout << ans;
}
