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;
long long int maxx, l, k, minn, w, d, n, m, x, z, y, a[100001], b, np, nl, c, p;
queue<int> s;
vector<long long int> v[100001];
bool mark[100001], markk[100001];
void clean(queue<int> &s) {
  while (!s.empty()) s.pop();
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    fill(a, a + 100001, 0);
    fill(mark, mark + 100001, 0);
    a[i] = 0;
    s.push(i);
    l = 1;
    while (s.size() != 0 && l != 0) {
      k = s.front();
      s.pop();
      z = v[k].size();
      for (int j = 0; j < z; j++) {
        if (mark[v[k][j]] == 0) {
          s.push(v[k][j]);
          mark[v[k][j]] = 1;
          a[v[k][j]] = a[k] + 1;
          if (a[v[k][j]] == 2 && v[k][j] > i) p++;
          if (a[v[k][j]] == 4) {
            l = 0;
          }
        }
      }
    }
    clean(s);
  }
  cout << p;
  return 0;
}
