#include <bits/stdc++.h>
struct data {
  int a, s, t;
  data() {}
  data(int a, int _s, int _t)
      : a(a), s(std::min(_s, _t)), t(std::max(_s, _t)) {}
};
int n, s, t;
int64_t ans;
std::bitset<5000> b[5000];
std::vector<data> hori, vert;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    static int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2)
      hori.emplace_back(x1, y1, y2);
    else
      vert.emplace_back(y1, x1, x2);
  }
  if (hori.size() > vert.size()) hori.swap(vert);
  s = (int)hori.size(), t = (int)vert.size();
  for (int i = 0; i < s; i++)
    for (int j = 0; j < t; j++) {
      auto &x = hori[i], &y = vert[j];
      if (x.s <= y.a && y.a <= x.t && y.s <= x.a && x.a <= y.t) b[i][j] = 1;
    }
  for (int i = 0; i < s; i++)
    for (int j = i + 1; j < s; j++) {
      int cnt = (int)(b[i] & b[j]).count();
      ans += cnt * (cnt - 1) / 2;
    }
  printf("%lld\n", ans);
  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>
struct data {
  int a, s, t;
  data() {}
  data(int a, int _s, int _t)
      : a(a), s(std::min(_s, _t)), t(std::max(_s, _t)) {}
};
int n, s, t;
int64_t ans;
std::bitset<5000> b[5000];
std::vector<data> hori, vert;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    static int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 == x2)
      hori.emplace_back(x1, y1, y2);
    else
      vert.emplace_back(y1, x1, x2);
  }
  if (hori.size() > vert.size()) hori.swap(vert);
  s = (int)hori.size(), t = (int)vert.size();
  for (int i = 0; i < s; i++)
    for (int j = 0; j < t; j++) {
      auto &x = hori[i], &y = vert[j];
      if (x.s <= y.a && y.a <= x.t && y.s <= x.a && x.a <= y.t) b[i][j] = 1;
    }
  for (int i = 0; i < s; i++)
    for (int j = i + 1; j < s; j++) {
      int cnt = (int)(b[i] & b[j]).count();
      ans += cnt * (cnt - 1) / 2;
    }
  printf("%lld\n", ans);
  return 0;
}

```