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 M = 1e9 + 7;
long long mpe(long long base, long long exponent, long long modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = result * base % modulus;
    exponent = exponent >> 1;
    base = base * base % modulus;
  }
  return result;
}
vector<int> vec[100005];
long long an = 0;
void dfs(int no, int p) {
  if (vec[no].size() == 1 && p != -1) {
    return;
  }
  long long cn = vec[no].size();
  an += (cn * (cn - 1)) / 2;
  for (auto i : vec[no]) {
    if (i == p) continue;
    dfs(i, no);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  long long n, m;
  cin >> n;
  m = n - 1;
  for (long long i = 0; i < m; ++i) {
    long long a, b;
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  dfs(1, -1);
  cout << an;
  return 0;
}
