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;
long long dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const long double pi = 3.141592653589;
const long long INF = 1e18;
const long long mod = 1000000007;
const long long maxn = 1e4 + 5;
template <typename T>
void printv(const vector<T>& v) {
  for (auto i : v) cout << i << ' ';
  cout << '\n';
}
template <typename T>
void printv(const vector<pair<T, T>>& v) {
  for (auto p : v) cout << "(" << p.first << "," << p.second << "),";
  cout << '\n';
}
long long n;
vector<long long> g[maxn];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans += (g[i].size() * (g[i].size() - 1) / 2);
  }
  cout << ans << '\n';
  return 0;
}
