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;
using graph_t = vector<list<int>>;
int dfs(int node, const graph_t& graph, vector<bool>& visited,
        vector<int>& lines) {
  visited[node] = true;
  int dist_1 = 0, dist_2 = 0;
  for (int adj : graph[node]) {
    if (!visited[adj]) {
      ++dist_1;
      dist_2 += dfs(adj, graph, visited, lines);
    }
  }
  lines[node] = dist_1 * (dist_1 - 1) / 2 + dist_2;
  return dist_1;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  graph_t graph(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  vector<bool> visited(n, false);
  vector<int> lines(n);
  dfs(0, graph, visited, lines);
  cout << accumulate(lines.begin(), lines.end(), 0);
}
