#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
struct Segm {
  int same;
  int low, high;
};
Segm vert[5000], horiz[5000];
int vert_size, horiz_size;
bitset<2500> arr[5000];
int arr_size;
bool Check(const Segm &one, const Segm &two) {
  if (one.same <= two.high && one.same >= two.low && two.same <= one.high &&
      two.same >= one.low)
    return true;
  return false;
}
signed main() {
  int n;
  scanf("%d", &n);
  int x1, y1, x2, y2;
  Segm tmp;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      tmp.same = x1;
      tmp.low = y1;
      tmp.high = y2;
      if (tmp.high < tmp.low) swap(tmp.high, tmp.low);
      vert[vert_size] = tmp;
      vert_size++;
    } else {
      tmp.same = y1;
      tmp.low = x1;
      tmp.high = x2;
      if (tmp.high < tmp.low) swap(tmp.high, tmp.low);
      horiz[horiz_size] = tmp;
      horiz_size++;
    }
  }
  if (vert_size > horiz_size) {
    swap(vert, horiz);
    swap(vert_size, horiz_size);
  }
  for (int i = 0; i < horiz_size; i++) {
    for (int j = 0; j < vert_size; j++) {
      arr[i][j] = Check(horiz[i], vert[j]);
    }
  }
  int count;
  long long ans = 0;
  for (int i = 0; i < horiz_size; i++) {
    for (int j = i + 1; j < horiz_size; j++) {
      count = (arr[i] & arr[j]).count();
      ans += count * (count - 1) / 2;
    }
  }
  printf("%lld", ans);
  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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
struct Segm {
  int same;
  int low, high;
};
Segm vert[5000], horiz[5000];
int vert_size, horiz_size;
bitset<2500> arr[5000];
int arr_size;
bool Check(const Segm &one, const Segm &two) {
  if (one.same <= two.high && one.same >= two.low && two.same <= one.high &&
      two.same >= one.low)
    return true;
  return false;
}
signed main() {
  int n;
  scanf("%d", &n);
  int x1, y1, x2, y2;
  Segm tmp;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      tmp.same = x1;
      tmp.low = y1;
      tmp.high = y2;
      if (tmp.high < tmp.low) swap(tmp.high, tmp.low);
      vert[vert_size] = tmp;
      vert_size++;
    } else {
      tmp.same = y1;
      tmp.low = x1;
      tmp.high = x2;
      if (tmp.high < tmp.low) swap(tmp.high, tmp.low);
      horiz[horiz_size] = tmp;
      horiz_size++;
    }
  }
  if (vert_size > horiz_size) {
    swap(vert, horiz);
    swap(vert_size, horiz_size);
  }
  for (int i = 0; i < horiz_size; i++) {
    for (int j = 0; j < vert_size; j++) {
      arr[i][j] = Check(horiz[i], vert[j]);
    }
  }
  int count;
  long long ans = 0;
  for (int i = 0; i < horiz_size; i++) {
    for (int j = i + 1; j < horiz_size; j++) {
      count = (arr[i] & arr[j]).count();
      ans += count * (count - 1) / 2;
    }
  }
  printf("%lld", ans);
  return 0;
}

```