#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 7;
vector<array<int, 3> > v1, v2;
bool intersect(const array<int, 3>& lhs, const array<int, 3>& rhs) {
  if (lhs[0] < rhs[1] || lhs[0] > rhs[2]) return 0;
  return rhs[0] >= lhs[1] && rhs[0] <= lhs[2];
}
bitset<N> sta[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int x1, x2, y1, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      if (y1 > y2) swap(y1, y2);
      v1.push_back({x1, y1, y2});
    } else {
      if (x1 > x2) swap(x1, x2);
      v2.push_back({y1, x1, x2});
    }
  }
  if (v1.size() > v2.size()) swap(v1, v2);
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  int l1 = v1.size(), l2 = v2.size();
  for (int i = 0; i < l1; i++)
    for (int j = 0; j < l2; j++)
      if (intersect(v1[i], v2[j])) sta[i].set(j);
  long long ans = 0;
  for (int i = 0; i < l1; i++) {
    for (int j = i + 1; j < l1; j++) {
      int t = (sta[i] & sta[j]).count();
      ans += t * (t - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int N = 5e3 + 7;
vector<array<int, 3> > v1, v2;
bool intersect(const array<int, 3>& lhs, const array<int, 3>& rhs) {
  if (lhs[0] < rhs[1] || lhs[0] > rhs[2]) return 0;
  return rhs[0] >= lhs[1] && rhs[0] <= lhs[2];
}
bitset<N> sta[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  int x1, x2, y1, y2;
  for (int i = 0; i < n; i++) {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      if (y1 > y2) swap(y1, y2);
      v1.push_back({x1, y1, y2});
    } else {
      if (x1 > x2) swap(x1, x2);
      v2.push_back({y1, x1, x2});
    }
  }
  if (v1.size() > v2.size()) swap(v1, v2);
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  int l1 = v1.size(), l2 = v2.size();
  for (int i = 0; i < l1; i++)
    for (int j = 0; j < l2; j++)
      if (intersect(v1[i], v2[j])) sta[i].set(j);
  long long ans = 0;
  for (int i = 0; i < l1; i++) {
    for (int j = i + 1; j < l1; j++) {
      int t = (sta[i] & sta[j]).count();
      ans += t * (t - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

```