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 int MAXN = 10000 + 5;
vector<int> G[MAXN];
long long int ans, dp[MAXN][10];
void dfs(int root, int f, int x) {
  for (int i = 0; i <= x; i++) dp[root][i] = 0;
  dp[root][0] = 1;
  for (int i = 0; i < G[root].size(); i++) {
    int son = G[root][i];
    if (son != f) {
      dfs(son, root, x);
      for (int j = 0; j < x; j++) ans += dp[son][j] * dp[root][x - j - 1];
      for (int j = 1; j <= x; j++) dp[root][j] += dp[son][j - 1];
    }
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    for (int i = 0; i <= n; i++) {
      G[i].clear();
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    ans = 0;
    dfs(1, 0, 2);
    printf("%I64d\n", ans);
  }
  return 0;
}
