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;
vector<int> q;
void bfs(map<int, vector<int>>& m, vector<int>& s, int& k, long long& ans,
         vector<bool>& v) {
  if (!v[q[k]]) {
    v[q[k]] = true;
    ans += (long long)(s[q[k]]) * (s[q[k]] - 1) / 2;
    for (int y : m[q[k]]) {
      if (!v[y]) {
        ans += --s[y];
        q.push_back(y);
      }
    }
    if (k + 1 < q.size()) {
      k++;
      bfs(m, s, k, ans, v);
    }
  }
}
int main() {
  int n, a, b, k = 0;
  long long ans = 0;
  cin >> n;
  map<int, vector<int>> m;
  vector<bool> v(n + 1, false);
  vector<int> s(n + 1);
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    m[a].push_back(b);
    m[b].push_back(a);
    s[a]++;
    s[b]++;
  }
  for (auto x : m) {
    if (!v[x.first]) {
      q.push_back(x.first);
      bfs(m, s, k, ans, v);
    }
  }
  cout << ans;
  return 0;
}
