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>
inline void cmax(T &x, T y) {
  x < y ? x = y : x;
}
template <typename T>
inline void cmin(T &x, T y) {
  y < x ? x = y : x;
}
char bf[1 << 21], *p1(bf), *p2(bf);
template <typename T>
inline void read(T &x) {
  char t(
      (p1 == p2 && (p1 = bf, p2 = bf + fread(bf, 1, 1 << 21, stdin), p1 == p2)
           ? EOF
           : *p1++)),
      flg(0);
  x = 0;
  for (; !isdigit(t);
       t = (p1 == p2 &&
                    (p1 = bf, p2 = bf + fread(bf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    flg = t == '-';
  for (; isdigit(t);
       t = (p1 == p2 &&
                    (p1 = bf, p2 = bf + fread(bf, 1, 1 << 21, stdin), p1 == p2)
                ? EOF
                : *p1++))
    x = x * 10 + (t & 15);
  flg ? x = -x : x;
}
clock_t t_bg, t_ed;
int N;
int hd[10005], nxt[10005 << 1], to[10005 << 1], tot;
inline void addedge(int x, int y) {
  nxt[++tot] = hd[x], hd[x] = tot, to[tot] = y;
}
long long ans;
int DFS(int u, int fa) {
  int c(0);
  for (int i(hd[u]), v(to[i]); i; v = to[i = nxt[i]])
    if (v != fa) {
      ++c;
      ans += DFS(v, u);
    }
  ans += c * (c - 1) / 2;
  return c;
}
int main() {
  t_bg = clock();
  read(N);
  int x, y;
  for (int i(2), I(N); i <= I; ++i)
    read(x), read(y), addedge(x, y), addedge(y, x);
  DFS(1, 0);
  cout << ans << endl;
  t_ed = clock();
  fprintf(stderr, "\n========info========\ntime : %.3f\n====================\n",
          (double)(t_ed - t_bg) / CLOCKS_PER_SEC);
  return 0;
}
