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;
int bfs(vector<vector<int> > vv) {
  int n = vv.size();
  int res = 0;
  bool c[n];
  int p[n];
  for (int i = 0; i < n; i++) {
    c[i] = false;
    p[i] = -1;
  }
  queue<int> q;
  c[0] = true;
  q.push(0);
  int x, y, d;
  while (!q.empty()) {
    x = q.front();
    q.pop();
    d = vv[x].size();
    res += x == 0 ? (d * (d - 1)) / 2 : ((d - 1) * (d - 2)) / 2;
    for (int j = 0; j < d; j++) {
      y = vv[x][j];
      if (!c[y]) {
        c[y] = true;
        p[y] = x;
        if (p[x] != -1) {
          res++;
        }
        q.push(y);
      }
    }
  }
  return res;
}
int main() {
  int n, a, b;
  cin >> n;
  vector<vector<int> > vv(n);
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    vv[a - 1].push_back(b - 1);
    vv[b - 1].push_back(a - 1);
  }
  cout << bfs(vv) << endl;
  return 0;
}
