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 INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 300007;
int n;
long long ans;
vector<int> ch[MAXN];
vector<int> dp[MAXN];
vector<int> tmp[MAXN];
void init();
void input();
void work();
void dfs(int now, int fa);
int getans(int now, vector<int> &cd);
void uni(vector<int> &a, int &va, vector<int> &b, int &vb);
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() {
  scanf("%d", &n);
  int u, v;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    ch[u].push_back(v), ch[v].push_back(u);
  }
}
void work() {
  dfs(1, -1);
  ans = static_cast<long long>(n) * n;
  vector<int> tt;
  getans(1, tt);
  cout << ans << endl;
}
void dfs(int now, int fa) {
  for (int i = 0; i < ch[now].size(); ++i) {
    if (ch[now][i] == fa) {
      ch[now].erase(ch[now].begin() + i);
      break;
    }
  }
  int nd = ch[now].size();
  for (int i = 0; i < nd; ++i) dfs(ch[now][i], now);
  dp[now].resize(nd + 1);
  for (int i = 1; i <= nd; ++i) tmp[i].clear();
  for (int i = 0; i < nd; ++i) {
    int v = ch[now][i];
    for (int j = 1; j <= ch[v].size(); ++j) {
      tmp[j].push_back(dp[v][j]);
    }
  }
  for (int i = 1; i <= nd; ++i) {
    if (tmp[i].size() >= i) {
      nth_element(tmp[i].begin(), tmp[i].end() - i, tmp[i].end());
      dp[now][i] = (*(tmp[i].end() - i)) + 1;
    } else
      dp[now][i] = 1;
  }
}
int getans(int now, vector<int> &cd) {
  vector<int> vd;
  int nd = ch[now].size();
  cd.resize(nd + 1);
  int ca = 0, va;
  for (int i = 1; i <= nd; ++i) cd[i] = dp[now][i], ca += cd[i];
  for (int i = 0; i < nd; ++i) {
    vd.clear();
    va = getans(ch[now][i], vd);
    uni(cd, ca, vd, va);
  }
  ans += ca;
  return ca;
}
void uni(vector<int> &a, int &va, vector<int> &b, int &vb) {
  if (a.size() < b.size()) {
    uni(b, vb, a, va);
    a.swap(b);
    swap(va, vb);
    return;
  }
  int bs = b.size();
  for (int i = 1; i < bs; ++i) {
    if (b[i] > a[i]) va += b[i] - a[i], a[i] = b[i];
  }
}
