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;
double sin_deg(double x) { return sin((x * acos(-1.0)) / 180.0); }
double cos_deg(double x) { return cos((x * acos(-1.0)) / 180.0); }
double tan_deg(double x) { return tan((x * acos(-1.0)) / 180.0); }
double asin_deg(double x) { return (asin(x) * 180) / acos(-1.0); }
double acos_deg(double x) { return (acos(x) * 180) / acos(-1.0); }
double atan_deg(double x) { return (atan(x) * 180) / acos(-1.0); }
long long int _xor(long long int a, long long int b) { return a ^ b; }
long long int _and(long long int a, long long int b) { return a & b; }
long long int _or(long long int a, long long int b) { return a | b; }
long long int _not(long long int a) { return ~a; }
vector<long long int> vec[1000000];
int main() {
  long long int a, b, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, w, x,
      y, z;
  long long int cnt = 0, ans = 0, l1, l2, l3, c1 = 0, c2 = 0, c3 = 0, mx = 0,
                mn = 0, f1, f2, f3, v1, v2, v3, tst, t1, t2, t3, t4;
  long long int prevsum = 0;
  string str, str1, str2;
  scanf("%lld", &n);
  long long int u[n - 1], v[n - 1];
  for (i = 0; i < n - 1; i++) {
    scanf("%lld", &u[i]);
    scanf("%lld", &v[i]);
    vec[u[i]].push_back(v[i]);
    vec[v[i]].push_back(u[i]);
  }
  long long int c = 0;
  for (i = 1; i <= n; i++) {
    c += (vec[i].size() * (vec[i].size() - 1)) / 2;
  }
  cout << c;
  return 0;
}
