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, m, t;
bool use[12000];
struct ac {
  int v, next;
} tab[12000], node[12000 * 2];
int cur;
int ans;
void insert(int x, int y) {
  node[m].v = y;
  node[m].next = tab[x].next;
  tab[x].next = m++;
}
void input() {
  int i;
  for (i = 0; i <= n; i++) {
    tab[i].next = -1;
    use[i] = false;
  }
  int x, y;
  for (i = 1; i < n; i++) {
    scanf("%d %d", &x, &y);
    insert(x - 1, y - 1);
    insert(y - 1, x - 1);
  }
}
void dfs(int u, int lay) {
  if (lay == 2) {
    ans++;
    return;
  }
  int rt = 1;
  int p = tab[u].next;
  while (p != -1) {
    int v = node[p].v;
    if (v != cur) {
      dfs(v, lay + 1);
    }
    p = node[p].next;
  }
}
void solve() {
  int i;
  m = t = 0;
  ans = 0;
  for (i = 0; i < n; i++) {
    cur = i;
    dfs(i, 0);
  }
  printf("%d\n", ans / 2);
}
int main() {
  while (scanf("%d", &n) != EOF) {
    input();
    solve();
  }
  return 0;
}
