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;
int n, ans = 0;
vector<vector<int> > g;
int dfs(int v, int p = -1) {
  int h_1 = int(g[v].size());
  if (p != -1) {
    h_1--;
  }
  for (int i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to == p) {
      continue;
    }
    ans += dfs(to, v);
  }
  ans += h_1 * (h_1 - 1) >> 1;
  return h_1;
}
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  g.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int from, to;
    scanf("%d%d", &from, &to);
    from--;
    to--;
    g[from].push_back(to);
    g[to].push_back(from);
  }
  dfs(0);
  printf("%d", ans);
  return 0;
}
