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;
const int N = 2e5 + 121;
int a[N], b[N], last[N], kol, ans, v, u, i, n;
inline void add(int v, int u) {
  a[++kol] = u;
  b[kol] = last[v];
  last[v] = kol;
}
void dfs(int v, int p, int len) {
  if (len == 2) {
    ++ans;
    return;
  }
  int x = last[v];
  while (x) {
    if (a[x] != p) dfs(a[x], v, len + 1);
    x = b[x];
  }
}
int main() {
  cin >> n;
  for (i = 1; i < n; ++i) {
    cin >> v >> u;
    add(v, u);
    add(u, v);
  }
  for (i = 1; i <= n; ++i) dfs(i, 0, 0);
  cout << ans / 2 << endl;
}
