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;
int n, i, j, a, b, S, parent[10005];
vector<int> g[10005];
long long answ;
void dfs(int v, int r) {
  if (r >= 2) answ++;
  long long z = 0;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == parent[v]) continue;
    parent[to] = v;
    z++;
    dfs(to, r + 1);
  }
  answ += z * (z - 1) / (2ll);
}
int main() {
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (i = 1; i <= n; i++)
    if (g[i].size() > 1) break;
  S = i == n + 1 ? 1 : i;
  dfs(S, 0);
  cout << answ << endl;
  return 0;
}
