#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 10;
const double pi = acos(-1.0);
const double eps = 1e-8;
struct node1 {
  int y, x1, x2;
} a[5050];
int tp1;
struct node2 {
  int x, y1, y2;
} b[5050];
int tp2;
bitset<5050> bs[5050];
int main() {
  int n;
  tp1 = tp2 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      b[++tp2].x = x1;
      if (y1 > y2) swap(y1, y2);
      b[tp2].y1 = y1;
      b[tp2].y2 = y2;
    } else {
      a[++tp1].y = y1;
      if (x1 > x2) swap(x1, x2);
      a[tp1].x1 = x1;
      a[tp1].x2 = x2;
    }
  }
  if (tp1 > tp2) {
    int tp = max(tp1, tp2);
    for (int i = 1; i <= tp; i++) {
      int tmp = a[i].y;
      a[i].y = b[i].x;
      b[i].x = tmp;
      tmp = a[i].x1;
      a[i].x1 = b[i].y1;
      b[i].y1 = tmp;
      tmp = a[i].x2;
      a[i].x2 = b[i].y2;
      b[i].y2 = tmp;
    }
    swap(tp1, tp2);
  }
  for (int i = 1; i <= tp1; i++) {
    for (int j = 1; j <= tp2; j++) {
      if (a[i].y >= b[j].y1 && a[i].y <= b[j].y2 && b[j].x >= a[i].x1 &&
          b[j].x <= a[i].x2) {
        bs[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i < tp1; i++) {
    if (bs[i].count() == 0) continue;
    for (int j = i + 1; j <= tp1; j++) {
      bitset<5050> bb = bs[i] & bs[j];
      int tmp = bb.count();
      ans += tmp * (tmp - 1) / 2;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 inf = 1e9 + 10;
const double pi = acos(-1.0);
const double eps = 1e-8;
struct node1 {
  int y, x1, x2;
} a[5050];
int tp1;
struct node2 {
  int x, y1, y2;
} b[5050];
int tp2;
bitset<5050> bs[5050];
int main() {
  int n;
  tp1 = tp2 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    if (x1 == x2) {
      b[++tp2].x = x1;
      if (y1 > y2) swap(y1, y2);
      b[tp2].y1 = y1;
      b[tp2].y2 = y2;
    } else {
      a[++tp1].y = y1;
      if (x1 > x2) swap(x1, x2);
      a[tp1].x1 = x1;
      a[tp1].x2 = x2;
    }
  }
  if (tp1 > tp2) {
    int tp = max(tp1, tp2);
    for (int i = 1; i <= tp; i++) {
      int tmp = a[i].y;
      a[i].y = b[i].x;
      b[i].x = tmp;
      tmp = a[i].x1;
      a[i].x1 = b[i].y1;
      b[i].y1 = tmp;
      tmp = a[i].x2;
      a[i].x2 = b[i].y2;
      b[i].y2 = tmp;
    }
    swap(tp1, tp2);
  }
  for (int i = 1; i <= tp1; i++) {
    for (int j = 1; j <= tp2; j++) {
      if (a[i].y >= b[j].y1 && a[i].y <= b[j].y2 && b[j].x >= a[i].x1 &&
          b[j].x <= a[i].x2) {
        bs[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i < tp1; i++) {
    if (bs[i].count() == 0) continue;
    for (int j = i + 1; j <= tp1; j++) {
      bitset<5050> bb = bs[i] & bs[j];
      int tmp = bb.count();
      ans += tmp * (tmp - 1) / 2;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```