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;
pair<int, int> sorted_pair(int x, int y) {
  if (x > y) {
    return pair<int, int>(y, x);
  } else {
    return pair<int, int>(x, y);
  }
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> graph(n + 1, vector<int>());
  set<pair<int, int>> pairs;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    queue<pair<int, int>> q;
    q.push(pair<int, int>(i, 0));
    vector<bool> visited(n + 1, false);
    while (!q.empty()) {
      int v = q.front().first;
      int lvl = q.front().second;
      visited[v] = true;
      if (lvl < 2) {
        for (int k = 0; k < graph[v].size(); k++) {
          if (!visited[graph[v][k]]) {
            q.push(pair<int, int>(graph[v][k], lvl + 1));
          }
        }
      } else if (lvl == 2) {
        pairs.insert(sorted_pair(i, v));
      }
      q.pop();
    }
  }
  cout << pairs.size() << endl;
  return 0;
}
