#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
bool ch(int a1, int a2, int a3, int a4) {
  if (a1 > a2) swap(a1, a2);
  if (a3 > a4) swap(a3, a4);
  return (a1 <= a3 && a2 >= a3) || (a3 <= a2 && a4 >= a2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> vv(n), h, v;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    vv[i] = {x1, y1, x2, y2, i};
    if (x1 == x2)
      v.push_back(vv[i]);
    else
      h.push_back(vv[i]);
  }
  if (h.size() > v.size()) swap(h, v);
  vector<bitset<5000>> vb(h.size());
  for (int i = 0; i < h.size(); i++) {
    for (auto v1 : v) {
      if (ch(h[i][0], h[i][2], v1[0], v1[2]) &&
          ch(h[i][1], h[i][3], v1[1], v1[3])) {
        vb[i][v1[4]] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < h.size(); i++) {
    for (int j = i + 1; j < h.size(); j++) {
      long long val = (vb[i] & vb[j]).count();
      ans += val * (val - 1) / 2;
    }
  }
  cout << 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 MOD = 1e9 + 7;
bool ch(int a1, int a2, int a3, int a4) {
  if (a1 > a2) swap(a1, a2);
  if (a3 > a4) swap(a3, a4);
  return (a1 <= a3 && a2 >= a3) || (a3 <= a2 && a4 >= a2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> vv(n), h, v;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    vv[i] = {x1, y1, x2, y2, i};
    if (x1 == x2)
      v.push_back(vv[i]);
    else
      h.push_back(vv[i]);
  }
  if (h.size() > v.size()) swap(h, v);
  vector<bitset<5000>> vb(h.size());
  for (int i = 0; i < h.size(); i++) {
    for (auto v1 : v) {
      if (ch(h[i][0], h[i][2], v1[0], v1[2]) &&
          ch(h[i][1], h[i][3], v1[1], v1[3])) {
        vb[i][v1[4]] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < h.size(); i++) {
    for (int j = i + 1; j < h.size(); j++) {
      long long val = (vb[i] & vb[j]).count();
      ans += val * (val - 1) / 2;
    }
  }
  cout << ans;
  return 0;
}

```