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 int mod = 1000000007;
long long int powmod(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
vector<int> g[10001];
int vis[10001];
vector<int> p;
long long int c = 0;
void dfs(int v) {
  if (vis[v]) return;
  vis[v] = 1;
  long long int x = g[v].size();
  c += x * (x - 1) / 2;
  for (int i = (0); i < (g[v].size()); i++) {
    dfs(g[v][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, m, flag = 1, s, f, t = 0, ans = 0, k, u, v;
    cin >> n;
    memset(vis, 0, sizeof(vis));
    for (int i = (0); i < (n - 1); i++) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(1);
    cout << c << endl;
  }
  return 0;
}
