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 = 10000 + 7;
const int M = 59;
const int mid = M / 2;
const int mod = 1e9 + 9;
const int inf = 1e6 + 3;
const long long linf = 1ll * inf * inf;
const double pi = acos(-1);
const double eps = 1e-7;
const double ep = 1e-5;
const int maxn = 1e5 + 7;
const double PI = acos(-1);
int n;
vector<int> adj[N];
void solve() {
  cin >> n;
  for (int i = (1); i < (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int ans = 0;
  for (int u = (1); u < (n + 1); u++) {
    int cur = (adj[u]).size() * ((adj[u]).size() - 1) / 2;
    ans += cur;
  }
  cout << ans;
}
int main() {
  int T = 1;
  for (int i = (1); i < (T + 1); i++) {
    solve();
  }
}
