#include <bits/stdc++.h>
using namespace std;
const int N = 5050, X = 2 * N;
int n, fen[X];
long long ans;
struct line {
  int x1, y1, x2, y2;
  void input() {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    x1 += N, y1 += N, x2 += N, y2 += N;
  }
};
vector<line> a, b;
void add(int x) {
  for (x; x < X; x += x & -x) fen[x]++;
}
int get(int dex) {
  int res = 0;
  for (; dex; dex ^= dex & -dex) res += fen[dex];
  return res;
}
int get(int l, int r) { return get(r) - get(l - 1); }
void clear() { memset(fen, 0, sizeof fen); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    line x;
    x.input();
    if (x.x1 == x.x2)
      b.push_back(x);
    else
      a.push_back(x);
  }
  sort(a.begin(), a.end(), [](line a, line b) { return a.y1 < b.y1; });
  sort(b.begin(), b.end(), [](line a, line b) { return a.y2 < b.y2; });
  if (a.empty() || b.empty()) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i < a.size(); i++) {
    clear();
    int dex = b.size() - 1;
    for (int j = a.size() - 1; ~j; j--)
      if (a[j].y1 > a[i].y1) {
        while (dex >= 0 && b[dex].y2 >= a[j].y1) {
          if ((b[dex].x1 < a[i].x1 || b[dex].x1 > a[i].x2) == false &&
              b[dex].y1 <= a[i].y1 && b[dex].y2 >= a[i].y2)
            add(b[dex].x1);
          dex--;
        }
        ans += 1LL * get(a[j].x1, a[j].x2) * (get(a[j].x1, a[j].x2) - 1) / 2;
      }
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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 N = 5050, X = 2 * N;
int n, fen[X];
long long ans;
struct line {
  int x1, y1, x2, y2;
  void input() {
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    x1 += N, y1 += N, x2 += N, y2 += N;
  }
};
vector<line> a, b;
void add(int x) {
  for (x; x < X; x += x & -x) fen[x]++;
}
int get(int dex) {
  int res = 0;
  for (; dex; dex ^= dex & -dex) res += fen[dex];
  return res;
}
int get(int l, int r) { return get(r) - get(l - 1); }
void clear() { memset(fen, 0, sizeof fen); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    line x;
    x.input();
    if (x.x1 == x.x2)
      b.push_back(x);
    else
      a.push_back(x);
  }
  sort(a.begin(), a.end(), [](line a, line b) { return a.y1 < b.y1; });
  sort(b.begin(), b.end(), [](line a, line b) { return a.y2 < b.y2; });
  if (a.empty() || b.empty()) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i < a.size(); i++) {
    clear();
    int dex = b.size() - 1;
    for (int j = a.size() - 1; ~j; j--)
      if (a[j].y1 > a[i].y1) {
        while (dex >= 0 && b[dex].y2 >= a[j].y1) {
          if ((b[dex].x1 < a[i].x1 || b[dex].x1 > a[i].x2) == false &&
              b[dex].y1 <= a[i].y1 && b[dex].y2 >= a[i].y2)
            add(b[dex].x1);
          dex--;
        }
        ans += 1LL * get(a[j].x1, a[j].x2) * (get(a[j].x1, a[j].x2) - 1) / 2;
      }
  }
  cout << ans;
  return 0;
}

```