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;
vector<int> g[10009];
int visit[10009];
int main() {
  int n, i, j, x, y;
  scanf("%d", &n);
  memset(visit, 0, sizeof(visit));
  for (i = 1; i <= n - 1; i++) {
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int count = 0;
  for (i = 1; i <= n; i++) {
    int size = g[i].size();
    for (j = 0; j < size; j++) {
      int v = g[i][j];
      int s = g[v].size();
      for (int k = 0; k < s; k++) {
        int u = g[v][k];
        if (!visit[u] && u != i) count++;
      }
    }
    visit[i] = 1;
  }
  printf("%d\n", count);
  return 0;
}
