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 GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const int MAXN = 1e5 + 9;
const int MOD = (int)(1e9 + 7);
const int LOG2MAXN = 17;
const int INF = 1e9;
int di[] = {0, 1, 0, -1};
int dj[] = {1, 0, -1, 0};
char ddir[] = {'R', 'U', 'L', 'D'};
int n, m;
vector<vector<int> > gg;
bool used[MAXN];
long long dd[MAXN];
long long ans;
void input() {
  cin >> n;
  gg.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    gg[u].push_back(v);
    gg[v].push_back(u);
    dd[v]++;
    dd[u]++;
  }
}
void dfs(int v, int p = -1) {
  used[v] = 1;
  if (p != -1) {
    long long x = dd[v] - 1;
    ans += (x * (x - 1)) / 2;
  } else
    ans += (dd[v] * (dd[v] - 1)) / 2;
  used[v] = 1;
  for (auto to : gg[v]) {
    if (!used[to]) {
      ans += dd[to] - 1;
      dfs(to, v);
    }
  }
}
void solve() {
  dfs(0);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  input();
  solve();
  return 0;
}
