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;
long long MOD = 1e9 + 7;
const long double E = 1e-10;
unsigned char ccc;
bool _minus = false;
template <typename T>
inline T sqr(T t) {
  return (t * t);
}
inline void read(long long &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') break;
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
}
inline void read(int &n) {
  n = 0;
  _minus = false;
  while (true) {
    ccc = getchar();
    if (ccc == ' ' || ccc == '\n') break;
    if (ccc == '-') {
      _minus = true;
      continue;
    }
    n = n * 10 + ccc - '0';
  }
  if (_minus) n *= -1;
}
char wwww[12];
int kkkk;
inline void write(long long y) {
  long long x = y;
  kkkk = 0;
  if (!x)
    ++kkkk, wwww[kkkk] = '0';
  else
    while (x) {
      ++kkkk;
      wwww[kkkk] = char(x % 10 + '0');
      x /= 10;
    }
  for (int i = kkkk; i >= 1; --i) putchar(wwww[i]);
  putchar(' ');
}
const int MAX = 1e5 + 1;
vector<int> vec[MAX];
bool used[MAX];
int dist[MAX];
int size = 0;
int _pos = 0;
int max_dist = 0;
long long ans = 0;
long long sum(long long n) {
  if (n < 0) return 0;
  return n * (n - 1) / 2;
}
void dfs(int pos, int d) {
  size++;
  dist[pos] = d;
  used[pos] = true;
  ans += sum((int)vec[pos].size());
  for (int i = 0; i < (int)vec[pos].size(); i++) {
    if (!used[vec[pos][i]]) {
      dfs(vec[pos][i], d + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  srand(time(NULL));
  cout.precision(5);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  dfs(1, 0);
  cout << ans << "\n";
}
