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 double PI = M_PI;
const int INF = -1 * 1e9;
const long long limit = -1 * 1e16;
const int MOD1 = 1e9 + 711;
const int MOD2 = 1e9 + 933;
const int MOD3 = 1e9 + 993;
const int maxn = 1e6 + 100;
int main() {
  cout.tie(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<vector<int>> g(n + 1, vector<int>());
  for (int i = 0; i < n - 1; ++i) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < g[i].size(); ++j) {
      sum += g[g[i][j]].size() - 1;
    }
  }
  cout << sum / 2;
  cerr << "\n\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}
