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>
#pragma warning(disable : 4996)
using namespace std;
const int Max = 10100;
vector<int> G[Max];
int n;
void input() {
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
}
struct ed {
  int f;
  int d;
  int i;
};
int bfs(int root) {
  queue<ed> que;
  ed temp, te;
  temp.d = 0;
  temp.f = -1;
  temp.i = root;
  que.push(temp);
  int ans = 0;
  while (!que.empty()) {
    temp = que.front();
    que.pop();
    for (int i = 0; i < (int)G[temp.i].size(); i++) {
      int v = G[temp.i][i];
      if (v != temp.f) {
        if (temp.d == 1) {
          ans++;
        } else {
          te.d = temp.d + 1;
          te.f = temp.i;
          te.i = v;
          que.push(te);
        }
      }
    }
  }
  return ans;
}
int main() {
  int ans = 0;
  scanf("%d", &n);
  input();
  for (int i = 1; i <= n; i++) {
    ans += bfs(i);
  }
  cout << ans / 2 << endl;
  cin >> n;
  return 0;
}
