Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies.

On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of n points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges).

To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!

Input

The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.

Output

Print one integer – the number of lifelines in the tree.

Examples

Input

4
1 2
1 3
1 4


Output

3

Input

5
1 2
2 3
3 4
3 5


Output

4

Note

In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
#include <bits/stdc++.h>
using namespace std;
template <class T>
void read(T& num) {
  char CH;
  bool F = false;
  for (CH = getchar(); CH < '0' || CH > '9'; F = CH == '-', CH = getchar())
    ;
  for (num = 0; CH >= '0' && CH <= '9';
       num = num * 10 + CH - '0', CH = getchar())
    ;
  F && (num = -num);
}
int stk[70], tp;
template <class T>
inline void print(T p) {
  if (!p) {
    puts("0");
    return;
  }
  while (p) stk[++tp] = p % 10, p /= 10;
  while (tp) putchar(stk[tp--] + '0');
  putchar('\n');
}
const long long mod = 1e9 + 7;
const double PI = acos(-1.0);
const long long inf = 1e18;
const int N = 2e5 + 10;
const int maxn = 1005;
const double eps = 1e-10;
int n, m, ind[N];
vector<int> ve[N];
queue<int> qu;
int main() {
  int u, v;
  read(n);
  for (int i = 1; i <= n - 1; i++) {
    read(u);
    read(v);
    ind[u]++;
    ind[v]++;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans + ind[i] * (ind[i] - 1) / 2;
  }
  cout << ans << "\n";
  return 0;
}
