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;
vector<int> vt[10005];
int mark[10005];
int sum = 0;
struct node {
  int v, w;
} M;
void bfs(int u) {
  queue<node> que;
  M.v = u;
  M.w = 0;
  que.push(M);
  while (!que.empty()) {
    node k = que.front();
    que.pop();
    if (k.w > 2) break;
    for (int i = 0; i < vt[k.v].size(); i++) {
      int y = vt[k.v][i];
      if (mark[y] == 0) {
        mark[y] = 1;
        if (k.w + 1 == 2) sum++;
        M.v = y;
        M.w = k.w + 1;
        que.push(M);
      }
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    vt[u].push_back(v);
    vt[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    memset(mark, 0, sizeof(mark));
    mark[i] = 1;
    bfs(i);
  }
  printf("%d\n", sum / 2);
}
