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;
using ll = long long;
const int N = 1e5 + 4;
vector<int> circle;
bool vis[N];
int parent[N];
int chk[N];
int color[N];
int tim[N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int cas = 0;
  int n, m, i, j, cnt = 0, cnt1 = 0, cnt2 = 0, even = 0, odd = 0, len, k, r, l,
                  z = 0, x = 0, y = 0, flag = 0, sum = 0;
  int a = 0, b = 0, c = 0, d = 0, p, q, ans = 0LL, rem, quot, zero = 0, fst = 0,
      null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
      mn = INT_MAX, mx = INT_MIN;
  char ch;
  int h1, h2, m1, m2, h;
  int velo1, velo2, ac1, ac2, tim;
  int node, edge, u, v, cost;
  int best, worst;
  double nd, ad, bd, xd;
  string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j < adj[i].size(); j++) {
      x = adj[i][j];
      ans += adj[x].size() - 1;
    }
  }
  cout << ans / 2 << endl;
  return 0;
}
