#include <bits/stdc++.h>
#pragma GCC target("avx2")
using namespace std;
inline bool Cross(pair<int, pair<int, int> > &a,
                  pair<int, pair<int, int> > &b) {
  return b.second.first <= a.first & a.first <= b.second.second &
         a.second.first <= b.first & b.first <= a.second.second;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > const_x, const_y;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) const_x.push_back({x1, {y1, y2}});
    if (y1 == y2) const_y.push_back({y1, {x1, x2}});
  }
  if ((int)const_x.size() > (int)const_y.size()) swap(const_x, const_y);
  vector<bitset<5000> > cross((int)const_x.size());
  for (int i = 0; i < (int)const_x.size(); i++)
    for (int j = 0; j < (int)const_y.size(); j++)
      cross[i][j] = Cross(const_x[i], const_y[j]);
  long long ans = 0;
  for (int i = 0; i < (int)const_x.size(); i++)
    for (int j = i + 1; j < (int)const_x.size(); j++) {
      int common_count = (cross[i] & cross[j]).count();
      ans += common_count * (common_count - 1) / 2;
    }
  cout << ans << endl;
}

### Prompt

Create a solution in cpp for the following problem:

There are n segments drawn on a plane; the i-th segment connects two points (x_{i, 1}, y_{i, 1}) and (x_{i, 2}, y_{i, 2}). Each segment is non-degenerate, and is either horizontal or vertical — formally, for every i ∈ [1, n] either x_{i, 1} = x_{i, 2} or y_{i, 1} = y_{i, 2} (but only one of these conditions holds). Only segments of different types may intersect: no pair of horizontal segments shares any common points, and no pair of vertical segments shares any common points.

We say that four segments having indices h_1, h_2, v_1 and v_2 such that h_1 < h_2 and v_1 < v_2 form a rectangle if the following conditions hold:

  * segments h_1 and h_2 are horizontal; 
  * segments v_1 and v_2 are vertical; 
  * segment h_1 intersects with segment v_1; 
  * segment h_2 intersects with segment v_1; 
  * segment h_1 intersects with segment v_2; 
  * segment h_2 intersects with segment v_2. 



Please calculate the number of ways to choose four segments so they form a rectangle. Note that the conditions h_1 < h_2 and v_1 < v_2 should hold.

Input

The first line contains one integer n (1 ≤ n ≤ 5000) — the number of segments.

Then n lines follow. The i-th line contains four integers x_{i, 1}, y_{i, 1}, x_{i, 2} and y_{i, 2} denoting the endpoints of the i-th segment. All coordinates of the endpoints are in the range [-5000, 5000].

It is guaranteed that each segment is non-degenerate and is either horizontal or vertical. Furthermore, if two segments share a common point, one of these segments is horizontal, and another one is vertical.

Output

Print one integer — the number of ways to choose four segments so they form a rectangle.

Examples

Input


7
-1 4 -1 -2
6 -1 -2 -1
-2 3 6 3
2 -2 2 4
4 -1 4 3
5 3 5 1
5 2 1 2


Output


7


Input


5
1 5 1 0
0 1 5 1
5 4 0 4
4 2 4 0
4 3 4 5


Output


0

Note

The following pictures represent sample cases:

<image> <image>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
using namespace std;
inline bool Cross(pair<int, pair<int, int> > &a,
                  pair<int, pair<int, int> > &b) {
  return b.second.first <= a.first & a.first <= b.second.second &
         a.second.first <= b.first & b.first <= a.second.second;
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > const_x, const_y;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) const_x.push_back({x1, {y1, y2}});
    if (y1 == y2) const_y.push_back({y1, {x1, x2}});
  }
  if ((int)const_x.size() > (int)const_y.size()) swap(const_x, const_y);
  vector<bitset<5000> > cross((int)const_x.size());
  for (int i = 0; i < (int)const_x.size(); i++)
    for (int j = 0; j < (int)const_y.size(); j++)
      cross[i][j] = Cross(const_x[i], const_y[j]);
  long long ans = 0;
  for (int i = 0; i < (int)const_x.size(); i++)
    for (int j = i + 1; j < (int)const_x.size(); j++) {
      int common_count = (cross[i] & cross[j]).count();
      ans += common_count * (common_count - 1) / 2;
    }
  cout << ans << endl;
}

```