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>
namespace FastIO {
template <typename T>
inline void readInteger(T& ret) {
  char ch = getchar();
  bool flag = false;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    flag = true;
    ch = getchar();
  }
  ret = 0;
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + (ch - '0');
    ch = getchar();
  }
  if (flag) ret = -ret;
}
inline int nextInt() {
  int x;
  readInteger(x);
  return x;
}
inline long long nextLong() {
  long long x;
  readInteger(x);
  return x;
}
template <typename T>
inline void outInteger(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) {
    outInteger(x / 10);
  }
  putchar(x % 10 + '0');
}
inline void newline() { putchar('\n'); }
}  // namespace FastIO
using namespace std;
using namespace FastIO;
int deg[11000];
int main() {
  int N, u, v;
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    scanf("%d%d", &u, &v);
    u--, v--;
    deg[u]++;
    deg[v]++;
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    ans += (long long)deg[i] * (deg[i] - 1) / 2;
  }
  printf("%I64d\n", ans);
  return 0;
}
