#include <bits/stdc++.h>
using namespace std;
long long n, x[2], y[2], res;
struct line {
  long long s, a, b;
};
vector<line> dir1, dir2;
bitset<5002> B[5001];
bool is_cut(line x, line y) {
  return x.a <= y.s && y.s <= x.b && y.a <= x.s && x.s <= y.b;
}
void Solve() {
  cin >> n;
  for (long long j = 1, _b = n; j <= _b; j++) {
    for (long long i = 0, _b = 2; i < _b; i++) cin >> x[i] >> y[i];
    if (x[0] == x[1])
      dir1.push_back({x[0], min(y[0], y[1]), max(y[0], y[1])});
    else
      dir2.push_back({y[0], min(x[0], x[1]), max(x[0], x[1])});
  }
  if (dir1.size() > dir2.size()) swap(dir1, dir2);
  for (long long i = 0, _b = dir1.size(); i < _b; i++)
    for (long long j = 0, _b = dir2.size(); j < _b; j++) {
      B[i].set(j, is_cut(dir1[i], dir2[j]));
    }
  for (long long i = 0, _b = dir1.size(); i < _b; i++)
    for (long long j = 0, _b = i; j < _b; j++) {
      long long tmp = (B[i] & B[j]).count();
      res += tmp * (tmp - 1) / 2;
    }
  cout << res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Test_numbers = 1;
  while (Test_numbers--) Solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long n, x[2], y[2], res;
struct line {
  long long s, a, b;
};
vector<line> dir1, dir2;
bitset<5002> B[5001];
bool is_cut(line x, line y) {
  return x.a <= y.s && y.s <= x.b && y.a <= x.s && x.s <= y.b;
}
void Solve() {
  cin >> n;
  for (long long j = 1, _b = n; j <= _b; j++) {
    for (long long i = 0, _b = 2; i < _b; i++) cin >> x[i] >> y[i];
    if (x[0] == x[1])
      dir1.push_back({x[0], min(y[0], y[1]), max(y[0], y[1])});
    else
      dir2.push_back({y[0], min(x[0], x[1]), max(x[0], x[1])});
  }
  if (dir1.size() > dir2.size()) swap(dir1, dir2);
  for (long long i = 0, _b = dir1.size(); i < _b; i++)
    for (long long j = 0, _b = dir2.size(); j < _b; j++) {
      B[i].set(j, is_cut(dir1[i], dir2[j]));
    }
  for (long long i = 0, _b = dir1.size(); i < _b; i++)
    for (long long j = 0, _b = i; j < _b; j++) {
      long long tmp = (B[i] & B[j]).count();
      res += tmp * (tmp - 1) / 2;
    }
  cout << res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Test_numbers = 1;
  while (Test_numbers--) Solve();
  return 0;
}

```