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>
int main() {
  int total = 0;
  int source = 0;
  int destination = 0;
  std::cin >> total;
  std::unordered_multimap<int, int> edges;
  while (std::cin >> source >> destination) {
    edges.emplace(source, destination);
    edges.emplace(destination, source);
  }
  int count = 0;
  for (int i = 1; i <= total; ++i) {
    auto eq = edges.equal_range(i);
    for (auto it = eq.first; it != eq.second; ++it) {
      auto eq2 = edges.equal_range(it->second);
      for (auto it2 = eq2.first; it2 != eq2.second; ++it2) {
        if (it2->second != it->first) ++count;
      }
    }
  }
  std::cout << count / 2;
}
