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;
template <typename T>
void read(T& x) {
  static char c;
  static int f;
  for (c = getchar(), f = 1; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -f;
  for (x = 0; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + (c & 15);
  x *= f;
}
template <typename T>
void write(T x) {
  static char q[65];
  int cnt = 0;
  if (x < 0) putchar('-'), x = -x;
  q[++cnt] = x % 10, x /= 10;
  while (x) q[++cnt] = x % 10, x /= 10;
  while (cnt) putchar(q[cnt--] + '0');
}
const int maxn = 10004;
int d[maxn];
int main() {
  int n;
  read(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    read(u), read(v);
    ++d[u], ++d[v];
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) ans += 1ll * d[i] * (d[i] - 1) / 2;
  write(ans), putchar('\n');
  return 0;
}
