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>
class tree {
 private:
  long long vertices;
  std::list<long long>* adj;
  long long* degree;

 public:
  tree(long long vertices) {
    this->vertices = vertices;
    adj = new std::list<long long>[vertices];
    degree = new long long[vertices];
    for (long long i = 0; i < vertices; i++) degree[i] = 0;
  }
  void addEdge(long long a, long long b) {
    degree[a - 1] = degree[a - 1] + 1;
    degree[b - 1] = degree[b - 1] + 1;
    adj[a - 1].push_back(b - 1);
    adj[b - 1].push_back(a - 1);
    return;
  }
  long long calculate_path() {
    long long i, count = 0;
    std::list<long long>::iterator it;
    for (i = 0; i < vertices; i++) {
      for (it = adj[i].begin(); it != adj[i].end(); it++)
        count = count + degree[*it] - 1;
    }
    return count / 2;
  }
};
int main() {
  long long n, a, b;
  std::cin >> n;
  tree mytree(n);
  for (long long i = 0; i < n - 1; i++) {
    std::cin >> a >> b;
    mytree.addEdge(a, b);
  }
  std::cout << mytree.calculate_path() << std::endl;
  return 0;
}
