#include <bits/stdc++.h>
using namespace std;
bool intr(pair<int, pair<int, int> >& a, pair<int, pair<int, int> >& b) {
  return (a.first >= b.second.first && a.first <= b.second.second &&
          b.first >= a.second.first && b.first <= a.second.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > ver, hor;
  for (int i = 0, a, b, c, d; i < n; ++i) {
    cin >> a >> b >> c >> d;
    if (b == d)
      hor.push_back({b, {min(a, c), max(a, c)}});
    else
      ver.push_back({a, {min(b, d), max(b, d)}});
  }
  if ((int)(ver).size() > (int)(hor).size()) swap(ver, hor);
  vector<bitset<5000> > btst((int)(ver).size());
  for (int i = 0; i < (int)(ver).size(); ++i)
    for (int j = 0; j < (int)(hor).size(); ++j)
      btst[i][j] = intr(ver[i], hor[j]);
  long long int ans = 0;
  for (int i = 0; i < (int)(ver).size(); ++i)
    for (int j = i + 1; j < (int)(ver).size(); ++j) {
      long long int c = (btst[i] & btst[j]).count();
      ans += (c * (c - 1)) / 2;
    }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a cpp solution to 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>
using namespace std;
bool intr(pair<int, pair<int, int> >& a, pair<int, pair<int, int> >& b) {
  return (a.first >= b.second.first && a.first <= b.second.second &&
          b.first >= a.second.first && b.first <= a.second.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > ver, hor;
  for (int i = 0, a, b, c, d; i < n; ++i) {
    cin >> a >> b >> c >> d;
    if (b == d)
      hor.push_back({b, {min(a, c), max(a, c)}});
    else
      ver.push_back({a, {min(b, d), max(b, d)}});
  }
  if ((int)(ver).size() > (int)(hor).size()) swap(ver, hor);
  vector<bitset<5000> > btst((int)(ver).size());
  for (int i = 0; i < (int)(ver).size(); ++i)
    for (int j = 0; j < (int)(hor).size(); ++j)
      btst[i][j] = intr(ver[i], hor[j]);
  long long int ans = 0;
  for (int i = 0; i < (int)(ver).size(); ++i)
    for (int j = i + 1; j < (int)(ver).size(); ++j) {
      long long int c = (btst[i] & btst[j]).count();
      ans += (c * (c - 1)) / 2;
    }
  cout << ans << endl;
  return 0;
}

```