#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f, N = 5e3 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n;
struct seg {
  int x1, y1, x2, y2;
};
vector<seg> v, h;
bitset<N> cnt[N];
bool check(seg a, seg b) {
  if (a.x1 == a.x2) swap(a, b);
  if (a.y1 >= b.y1 && a.y1 <= b.y2 && b.x1 >= a.x1 && b.x1 <= a.x2) return 1;
  return 0;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int a = read(), b = read(), c = read(), d = read();
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      v.push_back({a, b, c, d});
    else if (b == d)
      h.push_back({a, b, c, d});
  }
  if (h.size() > v.size()) swap(h, v);
  for (int i = 0; i < (int)h.size(); i++) {
    for (int j = 0; j < (int)v.size(); j++) {
      if (check(h[i], v[j])) cnt[i][j] = 1;
    }
  }
  long long ans = 0;
  for (int i = 0; i < (int)h.size(); i++) {
    for (int j = i + 1; j < (int)h.size(); j++) {
      long long tmp = (cnt[i] & cnt[j]).count();
      ;
      ans += tmp * (tmp - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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 = 0x3f3f3f3f, N = 5e3 + 5;
inline int read() {
  int sum = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  return sum * f;
}
int n;
struct seg {
  int x1, y1, x2, y2;
};
vector<seg> v, h;
bitset<N> cnt[N];
bool check(seg a, seg b) {
  if (a.x1 == a.x2) swap(a, b);
  if (a.y1 >= b.y1 && a.y1 <= b.y2 && b.x1 >= a.x1 && b.x1 <= a.x2) return 1;
  return 0;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int a = read(), b = read(), c = read(), d = read();
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      v.push_back({a, b, c, d});
    else if (b == d)
      h.push_back({a, b, c, d});
  }
  if (h.size() > v.size()) swap(h, v);
  for (int i = 0; i < (int)h.size(); i++) {
    for (int j = 0; j < (int)v.size(); j++) {
      if (check(h[i], v[j])) cnt[i][j] = 1;
    }
  }
  long long ans = 0;
  for (int i = 0; i < (int)h.size(); i++) {
    for (int j = i + 1; j < (int)h.size(); j++) {
      long long tmp = (cnt[i] & cnt[j]).count();
      ;
      ans += tmp * (tmp - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```