#include <bits/stdc++.h>
using namespace std;
struct V {
  int x, y1, y2;
  bool operator<(const V &other) const { return x < other.x; }
};
struct H {
  int x1, x2, y;
  bool crosses(const V &x) const {
    return x1 <= x.x && x.x <= x2 && x.y1 <= y && y <= x.y2;
  }
  bool operator<(const H &other) const { return x2 < other.x2; }
};
const int Mx = (int)1e4 + 9;
int vals[Mx + 99];
struct FT {
  int n;
  FT(const int n) : n(n) { memset(vals, 0, sizeof vals); }
  void update(int i, const int x) {
    for (++i; i <= n; i += i & -i) vals[i] += x;
  }
  int query(int i) const {
    assert(i >= -1 && i < n);
    int sum = 0;
    for (++i; i; i -= i & -i) sum += vals[i];
    return sum;
  }
  int query(int i, int j) const { return query(j) - query(i - 1); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n;
  cin >> n;
  vector<V> vert;
  vector<H> hori;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5000;
    x2 += 5000;
    y1 += 5000;
    y2 += 5000;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      vert.push_back({x1, y1, y2});
    else {
      assert(y1 == y2);
      hori.push_back({x1, x2, y1});
    }
  }
  sort(vert.begin(), vert.end());
  long long ans = 0;
  for (int i = (int)vert.size() - 1; i >= 0; --i) {
    vector<H> v;
    for (const H &x : hori)
      if (x.crosses(vert[i])) v.push_back(x);
    sort(v.begin(), v.end());
    FT ft(Mx);
    for (int j = (int)vert.size() - 1; i < j; --j) {
      while (!v.empty() && v.back().x2 >= vert[j].x) {
        ft.update(v.back().y, 1);
        v.pop_back();
      }
      const int y1 = max(vert[i].y1, vert[j].y1);
      const int y2 = min(vert[i].y2, vert[j].y2);
      if (y1 < y2) {
        const long long c = ft.query(y1, y2);
        ans += c * (c - 1) / 2;
      }
    }
  }
  cout << ans;
  return 0;
}

### 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;
struct V {
  int x, y1, y2;
  bool operator<(const V &other) const { return x < other.x; }
};
struct H {
  int x1, x2, y;
  bool crosses(const V &x) const {
    return x1 <= x.x && x.x <= x2 && x.y1 <= y && y <= x.y2;
  }
  bool operator<(const H &other) const { return x2 < other.x2; }
};
const int Mx = (int)1e4 + 9;
int vals[Mx + 99];
struct FT {
  int n;
  FT(const int n) : n(n) { memset(vals, 0, sizeof vals); }
  void update(int i, const int x) {
    for (++i; i <= n; i += i & -i) vals[i] += x;
  }
  int query(int i) const {
    assert(i >= -1 && i < n);
    int sum = 0;
    for (++i; i; i -= i & -i) sum += vals[i];
    return sum;
  }
  int query(int i, int j) const { return query(j) - query(i - 1); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int n;
  cin >> n;
  vector<V> vert;
  vector<H> hori;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5000;
    x2 += 5000;
    y1 += 5000;
    y2 += 5000;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      vert.push_back({x1, y1, y2});
    else {
      assert(y1 == y2);
      hori.push_back({x1, x2, y1});
    }
  }
  sort(vert.begin(), vert.end());
  long long ans = 0;
  for (int i = (int)vert.size() - 1; i >= 0; --i) {
    vector<H> v;
    for (const H &x : hori)
      if (x.crosses(vert[i])) v.push_back(x);
    sort(v.begin(), v.end());
    FT ft(Mx);
    for (int j = (int)vert.size() - 1; i < j; --j) {
      while (!v.empty() && v.back().x2 >= vert[j].x) {
        ft.update(v.back().y, 1);
        v.pop_back();
      }
      const int y1 = max(vert[i].y1, vert[j].y1);
      const int y2 = min(vert[i].y2, vert[j].y2);
      if (y1 < y2) {
        const long long c = ft.query(y1, y2);
        ans += c * (c - 1) / 2;
      }
    }
  }
  cout << ans;
  return 0;
}

```