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 choose(long long n, long long k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInv(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  long long res = (x % m + m) % m;
  return res;
}
long long mpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
vector<long long> adj[10005];
long long ans = 0;
void dfs(long long i, vector<long long> visited, long long dis) {
  visited[i] = true;
  if (dis == 2) {
    ans++;
    return;
  }
  for (int j = 0; j < adj[i].size(); j++) {
    if (!visited[adj[i][j]]) {
      dfs(adj[i][j], visited, dis + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (long long i = 1; i < n + 1; i++) {
    vector<long long> visited(n + 1, 0);
    dfs(i, visited, 0);
  }
  cout << ans / 2 << endl;
}
