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;
template <class T>
inline T bigmod(T b, T p, T m) {
  T ret;
  if (p == 0) return 1;
  if (p & 1) {
    ret = (bigmod(b, p / 2, m) % m);
    return ((b % m) * ret * ret) % m;
  } else {
    ret = (bigmod(b, p / 2, m) % m);
    return (ret * ret) % m;
  }
}
int dx4[] = {1, -1, 0, 0};
int dy4[] = {0, 0, 1, -1};
int dx6[] = {0, 0, 1, -1, 0, 0};
int dy6[] = {1, -1, 0, 0, 0, 0};
int dz6[] = {0, 0, 0, 0, 1, -1};
int dx8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dy8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dkx8[] = {-1, 1, -1, 1, -2, -2, 2, 2};
int dky8[] = {2, 2, -2, -2, 1, -1, 1, -1};
int tc = 1;
const double eps = 1e-9;
const double pi = acos(-1.0);
const long long int mx = 1e4;
const long long int mod = 1e9 + 7;
bool visited[mx + 5];
vector<int> edges[mx + 5];
int level[mx + 5];
int n, cnt;
void dfs(int node, int l) {
  int i, k;
  visited[node] = true;
  level[node] = l;
  if (level[node] == 2) cnt++;
  for (i = 0; i < edges[node].size(); i++) {
    k = edges[node][i];
    if (!visited[k] && l < 2) dfs(k, l + 1);
  }
}
int main() {
  int i, a, b;
  while (cin >> n) {
    for (i = 1; i < n; i++) {
      scanf("%d %d", &a, &b);
      edges[a].push_back(b);
      edges[b].push_back(a);
    }
    cnt = 0;
    for (i = 1; i <= n; i++) {
      memset(level, 0, sizeof(level));
      memset(visited, 0, sizeof(visited));
      dfs(i, 0);
    }
    cout << cnt / 2 << "\n";
  }
  return 0;
}
