#include <bits/stdc++.h>
using namespace std;
struct segm {
  int x1, y1, x2, y2;
};
int inter(segm a, segm b) {
  if (a.x1 <= b.x2 && a.x2 >= b.x1 && a.y1 <= b.y2 && a.y2 >= b.y1) return 1;
  return 0;
}
vector<segm> h, v;
bitset<5005> ok[5005];
int main() {
  int n, i, x1, y1, x2, y2, j, aux;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.push_back({x1, y1, x2, y2});
    else
      h.push_back({x1, y1, x2, y2});
  }
  if (v.size() > h.size()) swap(v, h);
  for (i = 0; i < v.size(); i++)
    for (j = 0; j < h.size(); j++) {
      int auxi = inter(v[i], h[j]);
      ok[i].set(j, auxi);
    }
  long long rasp = 0;
  for (i = 0; i < v.size(); i++)
    for (j = 0; j < i; j++) {
      aux = (ok[i] & ok[j]).count();
      rasp = rasp + 1LL * aux * (aux - 1) / 2;
    }
  printf("%lld\n", rasp);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 segm {
  int x1, y1, x2, y2;
};
int inter(segm a, segm b) {
  if (a.x1 <= b.x2 && a.x2 >= b.x1 && a.y1 <= b.y2 && a.y2 >= b.y1) return 1;
  return 0;
}
vector<segm> h, v;
bitset<5005> ok[5005];
int main() {
  int n, i, x1, y1, x2, y2, j, aux;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.push_back({x1, y1, x2, y2});
    else
      h.push_back({x1, y1, x2, y2});
  }
  if (v.size() > h.size()) swap(v, h);
  for (i = 0; i < v.size(); i++)
    for (j = 0; j < h.size(); j++) {
      int auxi = inter(v[i], h[j]);
      ok[i].set(j, auxi);
    }
  long long rasp = 0;
  for (i = 0; i < v.size(); i++)
    for (j = 0; j < i; j++) {
      aux = (ok[i] & ok[j]).count();
      rasp = rasp + 1LL * aux * (aux - 1) / 2;
    }
  printf("%lld\n", rasp);
  return 0;
}

```