#include <bits/stdc++.h>
using namespace std;
const int offset = 5001;
const int MAXN = 5005;
struct hSegment {
  int index;
  int y;
  int x1, x2;
  hSegment() {}
  hSegment(int x1, int x2, int y, int index) {
    this->y = y;
    this->x1 = x1;
    this->x2 = x2;
    this->index = index;
  }
};
struct vSegment {
  int index;
  int x;
  int y1, y2;
  vSegment() {}
  vSegment(int y1, int y2, int x, int index) {
    this->x = x;
    this->y1 = y1;
    this->y2 = y2;
    this->index = index;
  }
};
bool operator<(const vSegment A, const vSegment B) {
  if (A.y2 != B.y2) return A.y2 < B.y2;
  return A.index < B.index;
}
bool operator>(const vSegment A, const vSegment B) {
  if (A.y2 != B.y2) return A.y2 > B.y2;
  return A.index > B.index;
}
struct fenwickTree {
  int tree[MAXN * 2];
  fenwickTree() { memset(this->tree, 0, sizeof(this->tree)); }
  void update(int index, int value) {
    while (index < MAXN * 2) {
      this->tree[index] += value;
      index += index & (-index);
    }
  }
  int query(int index) {
    int sum = 0;
    while (index > 0) {
      sum += this->tree[index];
      index -= index & (-index);
    }
    return sum;
  }
  int query(int l, int r) {
    if (l > r) return 0;
    return this->query(r) - this->query(l - 1);
  }
};
int n;
set<int> allY;
vector<vSegment> allVertical;
vector<vSegment> x2Vertical[MAXN * 2];
vector<hSegment> y2Horizontal[MAXN * 2];
fenwickTree *T;
priority_queue<vSegment, vector<vSegment>, greater<vSegment>> q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long answer = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      allVertical.push_back(vSegment(y1, y2, x1, i));
      x2Vertical[x1 + offset].push_back(vSegment(y1, y2, x1, i));
    } else {
      allY.insert(y1);
      y2Horizontal[y1 + offset].push_back(hSegment(x1, x2, y1, i));
    }
  }
  T = new fenwickTree();
  for (auto it1 = allY.begin(); it1 != allY.end(); it1++) {
    while (q.empty() == false) {
      T->update(q.top().x + offset, -1);
      q.pop();
    }
    for (int i = 0; i < allVertical.size(); i++) {
      if (allVertical[i].y1 <= *it1) {
        T->update(allVertical[i].x + offset, 1);
        q.push(allVertical[i]);
      }
    }
    for (auto it2 = next(it1); it2 != allY.end(); it2++) {
      while (q.empty() == false && q.top().y2 < *it2) {
        T->update(q.top().x + offset, -1);
        q.pop();
      }
      for (hSegment curr1 : y2Horizontal[*it1 + offset]) {
        for (hSegment curr2 : y2Horizontal[*it2 + offset]) {
          int minX = max(curr1.x1, curr2.x1);
          int maxX = min(curr1.x2, curr2.x2);
          long long curr = T->query(minX + offset, maxX + offset);
          answer += curr * (curr - 1) / 2;
        }
      }
    }
  }
  cout << answer << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 offset = 5001;
const int MAXN = 5005;
struct hSegment {
  int index;
  int y;
  int x1, x2;
  hSegment() {}
  hSegment(int x1, int x2, int y, int index) {
    this->y = y;
    this->x1 = x1;
    this->x2 = x2;
    this->index = index;
  }
};
struct vSegment {
  int index;
  int x;
  int y1, y2;
  vSegment() {}
  vSegment(int y1, int y2, int x, int index) {
    this->x = x;
    this->y1 = y1;
    this->y2 = y2;
    this->index = index;
  }
};
bool operator<(const vSegment A, const vSegment B) {
  if (A.y2 != B.y2) return A.y2 < B.y2;
  return A.index < B.index;
}
bool operator>(const vSegment A, const vSegment B) {
  if (A.y2 != B.y2) return A.y2 > B.y2;
  return A.index > B.index;
}
struct fenwickTree {
  int tree[MAXN * 2];
  fenwickTree() { memset(this->tree, 0, sizeof(this->tree)); }
  void update(int index, int value) {
    while (index < MAXN * 2) {
      this->tree[index] += value;
      index += index & (-index);
    }
  }
  int query(int index) {
    int sum = 0;
    while (index > 0) {
      sum += this->tree[index];
      index -= index & (-index);
    }
    return sum;
  }
  int query(int l, int r) {
    if (l > r) return 0;
    return this->query(r) - this->query(l - 1);
  }
};
int n;
set<int> allY;
vector<vSegment> allVertical;
vector<vSegment> x2Vertical[MAXN * 2];
vector<hSegment> y2Horizontal[MAXN * 2];
fenwickTree *T;
priority_queue<vSegment, vector<vSegment>, greater<vSegment>> q;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long answer = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, x2, y1, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      allVertical.push_back(vSegment(y1, y2, x1, i));
      x2Vertical[x1 + offset].push_back(vSegment(y1, y2, x1, i));
    } else {
      allY.insert(y1);
      y2Horizontal[y1 + offset].push_back(hSegment(x1, x2, y1, i));
    }
  }
  T = new fenwickTree();
  for (auto it1 = allY.begin(); it1 != allY.end(); it1++) {
    while (q.empty() == false) {
      T->update(q.top().x + offset, -1);
      q.pop();
    }
    for (int i = 0; i < allVertical.size(); i++) {
      if (allVertical[i].y1 <= *it1) {
        T->update(allVertical[i].x + offset, 1);
        q.push(allVertical[i]);
      }
    }
    for (auto it2 = next(it1); it2 != allY.end(); it2++) {
      while (q.empty() == false && q.top().y2 < *it2) {
        T->update(q.top().x + offset, -1);
        q.pop();
      }
      for (hSegment curr1 : y2Horizontal[*it1 + offset]) {
        for (hSegment curr2 : y2Horizontal[*it2 + offset]) {
          int minX = max(curr1.x1, curr2.x1);
          int maxX = min(curr1.x2, curr2.x2);
          long long curr = T->query(minX + offset, maxX + offset);
          answer += curr * (curr - 1) / 2;
        }
      }
    }
  }
  cout << answer << '\n';
}

```