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>
const int MAXN = 300010;
int head[MAXN], nxt[MAXN << 1], to[MAXN << 1], tot;
void adde(int b, int e) {
  nxt[++tot] = head[b];
  to[head[b] = tot] = e;
  nxt[++tot] = head[e];
  to[head[e] = tot] = b;
}
std::vector<int> dp[MAXN];
int deg[MAXN];
std::priority_queue<int, std::vector<int>, std::greater<int> > q[MAXN];
long long ans, sm[MAXN];
void dfs(int u, int fa = 0) {
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) ++deg[u], dfs(to[i], u);
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa)
      for (int j = 0; j < deg[to[i]] && j < deg[u]; ++j)
        q[j].push(dp[to[i]][j]);
  for (int i = 0; i < deg[u]; ++i) {
    while (q[i].size() > i + 1) q[i].pop();
    int t = q[i].size() <= i ? 1 : q[i].top() + 1;
    while (!q[i].empty()) q[i].pop();
    dp[u].push_back(t);
    sm[u] += t;
  }
}
void dfs2(int u, int fa = 0) {
  for (int i = head[u]; i; i = nxt[i])
    if (to[i] != fa) {
      dfs2(to[i], u);
      if (dp[to[i]].size() > dp[u].size()) {
        dp[u].swap(dp[to[i]]);
        std::swap(sm[u], sm[to[i]]);
      }
      const int S = dp[to[i]].size();
      for (int j = 0; j != S; ++j)
        if (dp[to[i]][j] > dp[u][j]) {
          sm[u] -= dp[u][j];
          sm[u] += dp[u][j] = dp[to[i]][j];
        }
    }
  ans += sm[u];
}
int n;
int main() {
  std::ios_base::sync_with_stdio(false), std::cin.tie(0);
  std::cin >> n;
  for (int i = 1, t1, t2; i < n; ++i) std::cin >> t1 >> t2, adde(t1, t2);
  ans = (long long)n * n;
  dfs(1);
  dfs2(1);
  std::cout << ans << std::endl;
  return 0;
}
