#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000 + 10;
const int INF = 0x3f3f3f3f;
struct node {
  int x1, y1, x2, y2;
} R[5000 + 10], C[5000 + 10];
bitset<5000 + 10> p1[5000 + 10], p2[5000 + 10];
bool judge(node i, node j) {
  if (i.y2 >= j.y1 && j.y1 >= i.y1 && j.x2 >= i.x1 && i.x1 >= j.x1) return true;
  return false;
}
int main() {
  int n, a, b, c, d;
  int sum1 = 0, sum2 = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    if (a == c) {
      R[sum1].x1 = a;
      R[sum1].y1 = min(b, d);
      R[sum1].x2 = c;
      R[sum1++].y2 = max(b, d);
    } else {
      C[sum2].x1 = min(a, c);
      C[sum2].y1 = b;
      C[sum2].x2 = max(a, c);
      C[sum2++].y2 = d;
    }
  }
  for (int i = 0; i < sum1; i++) {
    for (int j = 0; j < sum2; j++) {
      if (judge(R[i], C[j])) {
        p1[i][j] = 1;
        p2[j][i] = 1;
      }
    }
  }
  long long res = 0, sum;
  int N = min(sum1, sum2);
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (sum1 < sum2)
        sum = (p1[i] & p1[j]).count();
      else
        sum = (p2[i] & p2[j]).count();
      res += sum * (sum - 1) / 2;
    }
  }
  printf("%lld\n", res);
  return 0;
}

### Prompt

Please formulate 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;
const int maxn = 1000000 + 10;
const int INF = 0x3f3f3f3f;
struct node {
  int x1, y1, x2, y2;
} R[5000 + 10], C[5000 + 10];
bitset<5000 + 10> p1[5000 + 10], p2[5000 + 10];
bool judge(node i, node j) {
  if (i.y2 >= j.y1 && j.y1 >= i.y1 && j.x2 >= i.x1 && i.x1 >= j.x1) return true;
  return false;
}
int main() {
  int n, a, b, c, d;
  int sum1 = 0, sum2 = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    if (a == c) {
      R[sum1].x1 = a;
      R[sum1].y1 = min(b, d);
      R[sum1].x2 = c;
      R[sum1++].y2 = max(b, d);
    } else {
      C[sum2].x1 = min(a, c);
      C[sum2].y1 = b;
      C[sum2].x2 = max(a, c);
      C[sum2++].y2 = d;
    }
  }
  for (int i = 0; i < sum1; i++) {
    for (int j = 0; j < sum2; j++) {
      if (judge(R[i], C[j])) {
        p1[i][j] = 1;
        p2[j][i] = 1;
      }
    }
  }
  long long res = 0, sum;
  int N = min(sum1, sum2);
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (sum1 < sum2)
        sum = (p1[i] & p1[j]).count();
      else
        sum = (p2[i] & p2[j]).count();
      res += sum * (sum - 1) / 2;
    }
  }
  printf("%lld\n", res);
  return 0;
}

```