#include <bits/stdc++.h>
using namespace std;
class Solution {
 public:
  long long solve(vector<vector<int>>& segs) {
    vector<int> H;
    vector<int> V;
    int n = segs.size();
    for (int i = 0; i < n; ++i) {
      if (segs[i][0] == segs[i][2]) {
        if (segs[i][1] > segs[i][3]) {
          swap(segs[i][1], segs[i][3]);
        }
        V.push_back(i);
      } else {
        if (segs[i][0] > segs[i][2]) {
          swap(segs[i][0], segs[i][2]);
        }
        H.push_back(i);
      }
    }
    n = H.size();
    int m = V.size();
    if (n > m) {
      swap(n, m);
      swap(H, V);
    }
    auto intersect = [&](vector<int>& a, vector<int>& b) {
      if (a[0] == a[2]) {
        return b[0] <= a[0] && a[0] <= b[2] && a[1] <= b[1] && b[1] <= a[3];
      } else {
        return a[0] <= b[0] && b[0] <= a[2] && b[1] <= a[1] && a[1] <= b[3];
      }
    };
    vector<bitset<5010>> bits(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (intersect(segs[H[i]], segs[V[j]])) {
          bits[i].set(j);
        }
      }
    }
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        int cnt = (bits[i] & bits[j]).count();
        res += cnt * (cnt - 1) / 2;
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n;
  cin >> n;
  vector<vector<int>> segs;
  segs.reserve(n);
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    segs.emplace_back(vector<int>{x1, y1, x2, y2});
  }
  cout << sol.solve(segs) << '\n';
  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;
class Solution {
 public:
  long long solve(vector<vector<int>>& segs) {
    vector<int> H;
    vector<int> V;
    int n = segs.size();
    for (int i = 0; i < n; ++i) {
      if (segs[i][0] == segs[i][2]) {
        if (segs[i][1] > segs[i][3]) {
          swap(segs[i][1], segs[i][3]);
        }
        V.push_back(i);
      } else {
        if (segs[i][0] > segs[i][2]) {
          swap(segs[i][0], segs[i][2]);
        }
        H.push_back(i);
      }
    }
    n = H.size();
    int m = V.size();
    if (n > m) {
      swap(n, m);
      swap(H, V);
    }
    auto intersect = [&](vector<int>& a, vector<int>& b) {
      if (a[0] == a[2]) {
        return b[0] <= a[0] && a[0] <= b[2] && a[1] <= b[1] && b[1] <= a[3];
      } else {
        return a[0] <= b[0] && b[0] <= a[2] && b[1] <= a[1] && a[1] <= b[3];
      }
    };
    vector<bitset<5010>> bits(n);
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (intersect(segs[H[i]], segs[V[j]])) {
          bits[i].set(j);
        }
      }
    }
    long long res = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        int cnt = (bits[i] & bits[j]).count();
        res += cnt * (cnt - 1) / 2;
      }
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution sol;
  int n;
  cin >> n;
  vector<vector<int>> segs;
  segs.reserve(n);
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    segs.emplace_back(vector<int>{x1, y1, x2, y2});
  }
  cout << sol.solve(segs) << '\n';
  return 0;
}

```