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 = 3.1415926536;
struct Node {
  vector<int> to;
  int l;
  bool was;
  Node() {
    l = 0;
    was = false;
  }
};
vector<Node> g;
int s = 0;
void BFS(int v) {
  queue<pair<int, int>> q;
  q.push(make_pair(v, 0));
  while (!q.empty()) {
    int a, b;
    a = q.front().first;
    b = q.front().second;
    q.pop();
    if (g[a].was) continue;
    g[a].was = true;
    if (b > 2) continue;
    if (b == 2) {
      ++s;
    }
    for (int i = 0; i < g[a].to.size(); ++i) {
      q.push(make_pair(g[a].to[i], b + 1));
    }
  }
}
int main() {
  int n;
  cin >> n;
  g.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].to.push_back(b);
    g[b].to.push_back(a);
  }
  for (int i = 0; i < n; ++i) {
    BFS(i);
    for (int i = 0; i < n; ++i) {
      g[i].was = false;
    }
  }
  cout << s / 2 << endl;
  return 0;
}
