#include <bits/stdc++.h>
using namespace std;
const int N = 5005;
struct node {
  int x, y1, y2;
  bool operator<(const node &_) const { return x < _.x; }
} A[N];
struct Node {
  int y, x1, x2;
  bool operator<(const Node &_) const { return x2 > _.x2; }
} B[N];
int C[N << 1], nx, ny, ans[N], tol, n;
void Add(int x, int y) {
  x += 5001;
  while (x <= 10001) C[x] += y, x += x & -x;
}
int sum(int x) {
  int ret = 0;
  for (x += 5001; x; x ^= x & -x) ret += C[x];
  return ret;
}
int main() {
  int x1, y1, x2, y2;
  scanf("%d", &n);
  for (int i = 1, i_ = n; i <= i_; 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)
      A[++nx] = (node){x1, y1, y2};
    else
      B[++ny] = (Node){y1, x1, x2};
  }
  sort(A + 1, A + 1 + nx);
  sort(B + 1, B + 1 + ny);
  long long res = 0, i;
  for (int l = 1, l_ = nx; l <= l_; l++) {
    tol = 0, i = 1;
    for (int r = nx, r_ = l + 1; r >= r_; r--) {
      int x = min(A[l].y2, A[r].y2), y = max(A[l].y1, A[r].y1);
      if (y >= x) continue;
      while (i <= ny && B[i].x2 >= A[r].x) {
        if (B[i].x1 <= A[l].x) {
          Add(B[i].y, 1);
          ans[++tol] = B[i].y;
        }
        i++;
      }
      int ct = sum(x) - sum(y - 1);
      res += (long long)ct * (ct - 1) >> 1;
    }
    for (int i = 1, i_ = tol; i <= i_; i++) Add(ans[i], -1);
  }
  cout << res;
  return 0;
}

### 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;
const int N = 5005;
struct node {
  int x, y1, y2;
  bool operator<(const node &_) const { return x < _.x; }
} A[N];
struct Node {
  int y, x1, x2;
  bool operator<(const Node &_) const { return x2 > _.x2; }
} B[N];
int C[N << 1], nx, ny, ans[N], tol, n;
void Add(int x, int y) {
  x += 5001;
  while (x <= 10001) C[x] += y, x += x & -x;
}
int sum(int x) {
  int ret = 0;
  for (x += 5001; x; x ^= x & -x) ret += C[x];
  return ret;
}
int main() {
  int x1, y1, x2, y2;
  scanf("%d", &n);
  for (int i = 1, i_ = n; i <= i_; 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)
      A[++nx] = (node){x1, y1, y2};
    else
      B[++ny] = (Node){y1, x1, x2};
  }
  sort(A + 1, A + 1 + nx);
  sort(B + 1, B + 1 + ny);
  long long res = 0, i;
  for (int l = 1, l_ = nx; l <= l_; l++) {
    tol = 0, i = 1;
    for (int r = nx, r_ = l + 1; r >= r_; r--) {
      int x = min(A[l].y2, A[r].y2), y = max(A[l].y1, A[r].y1);
      if (y >= x) continue;
      while (i <= ny && B[i].x2 >= A[r].x) {
        if (B[i].x1 <= A[l].x) {
          Add(B[i].y, 1);
          ans[++tol] = B[i].y;
        }
        i++;
      }
      int ct = sum(x) - sum(y - 1);
      res += (long long)ct * (ct - 1) >> 1;
    }
    for (int i = 1, i_ = tol; i <= i_; i++) Add(ans[i], -1);
  }
  cout << res;
  return 0;
}

```