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;
void Read(int &x) {
  static char c;
  bool f(0);
  while (c = getchar(), c != EOF) {
    if (c == '-')
      f = 1;
    else if (c >= '0' && c <= '9') {
      x = c - '0';
      while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
      ungetc(c, stdin);
      if (f) x = -x;
      return;
    }
  }
}
int n, ans, son[10000 + 10];
struct node {
  int v;
  node *next;
} * adj[10000 + 10], edge[10000 * 2 + 10], *ecnt = edge;
inline void addedge(int u, int v) {
  node *p = ++ecnt;
  p->v = v;
  p->next = adj[u];
  adj[u] = p;
}
void dfs(int u, int fa) {
  for (node *p = adj[u]; p; p = p->next) {
    if (p->v != fa) {
      dfs(p->v, u);
      ans += son[p->v];
      son[u]++;
    }
  }
  ans += son[u] * (son[u] - 1) / 2;
}
void read() {
  Read(n);
  int i, u, v;
  for (i = 1; i < n; i++) {
    Read(u), Read(v);
    addedge(u, v);
    addedge(v, u);
  }
}
int main() {
  read();
  dfs(1, 0);
  printf("%d\n", ans);
}
