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 double EPS = 1e-5;
const int MOD = int(1e9) + 7;
const int MX = int(1e5) + 1, OO = int(1e9);
vector<vector<int> > Adj;
vector<int> vis;
int vs = 1;
long long BFS(int i) {
  queue<int> q;
  q.push(i);
  vis[i] = vs;
  long long cnt = 0;
  int lv = 0;
  while (int(q.size()) && lv < 2) {
    int sz = int(q.size());
    while (sz--) {
      int cur = q.front();
      q.pop();
      for (int ch : Adj[cur]) {
        if (vis[ch] != vs) {
          vis[ch] = vs;
          q.push(ch);
          if (lv == 1) ++cnt;
        }
      }
    }
    ++lv;
  }
  return cnt;
}
void solve() {
  int n;
  scanf("%d", &n);
  Adj.resize(n + 1);
  vis.assign(n + 1, 0);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    Adj[a].push_back(b);
    Adj[b].push_back(a);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ++vs;
    ans += BFS(i);
  }
  printf("%lld\n", ans / 2);
}
int main() { solve(); }
