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 = 0x3f3f3f3f, oo = inf;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
void writeln(long long x) {
  write(x);
  puts("");
}
const long long maxn = 3e5 + 233;
struct Edge {
  long long to, nxt;
  Edge() {}
  Edge(long long to, long long nxt) : to(to), nxt(nxt) {}
} edge[maxn * 2];
long long first[maxn], nume;
long long deep[maxn], dp[maxn][20], n;
long long ans = 0;
void Addedge(long long a, long long b) {
  edge[nume] = Edge(b, first[a]);
  first[a] = nume++;
}
bool cmp(long long a, long long b) { return a > b; }
long long q[maxn];
void dfs(long long u, long long fa) {
  for (long long e = first[u]; ~e; e = edge[e].nxt) {
    long long v = edge[e].to;
    if (v == fa) continue;
    dfs(v, u);
    deep[u] = max(deep[u], deep[v]);
  }
  deep[u]++;
  ans += deep[u];
  dp[u][1] = n;
  for (long long i = 2; i < 20; i++) {
    long long tot = 0;
    for (long long e = first[u]; ~e; e = edge[e].nxt) {
      long long v = edge[e].to;
      if (v == fa) continue;
      if (dp[v][i - 1] > 1) q[++tot] = dp[v][i - 1];
    }
    sort(q + 1, q + 1 + tot, cmp);
    for (long long j = tot; j >= 2; j--)
      if (q[j] >= j) {
        dp[u][i] = j;
        break;
      }
  }
}
void pushup(long long u, long long fa) {
  for (long long e = first[u]; ~e; e = edge[e].nxt) {
    long long v = edge[e].to;
    if (v == fa) continue;
    pushup(v, u);
    for (long long i = 0; i < 20; i++) {
      dp[u][i] = max(dp[u][i], dp[v][i]);
    }
  }
  for (long long i = 1; i < 19; i++) {
    ans = ans + i * (dp[u][i] - dp[u][i + 1]);
    if (dp[u][i + 1] == 0) {
      ans -= i;
      break;
    }
  }
}
signed main() {
  n = read();
  memset(first, -1, sizeof(first));
  nume = 0;
  for (register long long i = (1); i < (n); ++i) {
    long long a = read(), b = read();
    Addedge(a, b);
    Addedge(b, a);
  }
  dfs(1, 0);
  pushup(1, 0);
  writeln(ans);
  return 0;
}
