#include <bits/stdc++.h>
using namespace std;
struct graw {
  long long a;
  long long b;
  long long c;
};
vector<graw> vert;
vector<graw> hor;
bool check(graw a, graw b) {
  return b.b <= a.a && a.a <= b.c && a.b <= b.a && a.c >= b.a;
}
bitset<5005> xi[5505];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      vert.push_back({a, b, d});
    } else {
      hor.push_back({b, a, c});
    }
  }
  if (hor.size() > vert.size()) {
    swap(hor, vert);
  }
  for (long long i = 0; i < hor.size(); i++) {
    for (long long j = 0; j < vert.size(); j++) {
      if (check(hor[i], vert[j])) xi[i].set(j);
    }
  }
  long long ans = 0;
  for (long long i = 0; i < hor.size(); i++) {
    for (long long j = i + 1; j < hor.size(); j++) {
      long long k = (xi[i] & xi[j]).count();
      k = k * (k - 1) / 2;
      ans += k;
    }
  }
  cout << ans;
}

### Prompt

Generate 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;
struct graw {
  long long a;
  long long b;
  long long c;
};
vector<graw> vert;
vector<graw> hor;
bool check(graw a, graw b) {
  return b.b <= a.a && a.a <= b.c && a.b <= b.a && a.c >= b.a;
}
bitset<5005> xi[5505];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      vert.push_back({a, b, d});
    } else {
      hor.push_back({b, a, c});
    }
  }
  if (hor.size() > vert.size()) {
    swap(hor, vert);
  }
  for (long long i = 0; i < hor.size(); i++) {
    for (long long j = 0; j < vert.size(); j++) {
      if (check(hor[i], vert[j])) xi[i].set(j);
    }
  }
  long long ans = 0;
  for (long long i = 0; i < hor.size(); i++) {
    for (long long j = i + 1; j < hor.size(); j++) {
      long long k = (xi[i] & xi[j]).count();
      k = k * (k - 1) / 2;
      ans += k;
    }
  }
  cout << ans;
}

```