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;
int const N = 1e4 + 10;
int n;
vector<int> adj[N], dag[N];
void dfs0(int v, int par) {
  if (par != -1) dag[par].push_back(v);
  for (int i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u != par) dfs0(u, v);
  }
}
int dp[N], sum[N];
void dfs1(int v) {
  sum[v] = (int)(dag[v].size());
  for (int i = 0; i < dag[v].size(); i++) {
    int u = dag[v][i];
    dfs1(u);
    dp[v] += (sum[u] + dp[u]);
  }
  if (sum[v] > 1) dp[v] += (sum[v] * (sum[v] - 1) / 2);
}
void solve() {
  dfs0(1, -1);
  dfs1(1);
  cout << dp[1] << '\n';
}
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  solve();
  return 0;
}
