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;
const long long N = 200005, INF = 1000000000000000000;
long long power(long long a, long long b) {
  if (a == 0) return 0;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    b >>= 1;
    a = (a * a);
  }
  return res;
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(std::numeric_limits<double>::max_digits10);
  ;
  long long n, ans = 0;
  cin >> n;
  vector<long long> v[N];
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    long long l = v[i].size();
    for (long long j = 0; j < l; j++) ans += (v[v[i][j]].size() - 1);
  }
  cout << ans / 2;
}
