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;
inline long toInt(string s) {
  long v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
long long pw(long long x, long long y) {
  long long ans = 1;
  for (long long i = 0; i < y; i++) {
    ans = ans * x;
  }
  return ans;
}
long long factorial(long long x) {
  long long ans = 1;
  for (int i = 1; i < x + 1; i++) {
    ans = ans * i;
  }
  return ans;
}
long long combination(long long x, long long y) {
  return factorial(x) / factorial(y) / factorial(x - y);
}
long long permutation(long long x, long long y) {
  return factorial(x) / factorial(x - y);
}
string flip(string s) {
  string ss;
  for (int i = 0; i < s.size(); i++) {
    ss = toString(s[i]) + ss;
  }
  return ss;
}
long long mod(long long a, long long b) {
  if (a % b == 0) {
    return 0;
  } else {
    return a % b;
  }
}
int main() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  long long a, b;
  for (long long i = 0; i < n - 1; i++) {
    cin >> a >> b;
    v[a - 1]++;
    v[b - 1]++;
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    if (v[i] >= 2) {
      ans += combination(v[i], 2);
    }
  }
  cout << ans << endl;
  return 0;
}
