#include <bits/stdc++.h>
using namespace std;
int T;
struct Line {
  int x, y, xx, yy;
  Line(int X = 0, int Y = 0, int XX = 0, int YY = 0)
      : x(X), y(Y), xx(XX), yy(YY) {}
};
int n, m;
Line a[5005], b[5005];
bool cmp1(Line A, Line B) { return A.y < B.y; }
bool cmp2(Line A, Line B) { return A.x < B.x; }
int c[10005];
void add(int x, int v) {
  for (; x <= 10003; x += x & (-x)) c[x] += v;
}
int get(int x) {
  int ans = 0;
  for (; x; x -= x & (-x)) ans += c[x];
  return ans;
}
struct node {
  int y, pos;
  node(int Y = 0, int POS = 0) : y(Y), pos(POS) {}
};
bool operator<(node A, node B) { return A.y > B.y; }
int main() {
  scanf("%d", &T);
  for (int i = 1; i <= T; ++i) {
    int x, y, xx, yy;
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    if (x > xx) swap(x, xx);
    if (y > yy) swap(y, yy);
    x += 5002, y += 5002, xx += 5002, yy += 5002;
    if (y == yy)
      a[++n] = Line(x, y, xx, yy);
    else
      b[++m] = Line(x, y, xx, yy);
  }
  sort(a + 1, a + n + 1, cmp1);
  sort(b + 1, b + m + 1, cmp2);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    priority_queue<node> pq;
    memset(c, 0, sizeof(c));
    for (int k = 1; k <= m; ++k)
      if (a[i].x <= b[k].x && b[k].x <= a[i].xx && b[k].y <= a[i].y &&
          a[i].y <= b[k].yy) {
        add(b[k].x, 1);
        pq.push(node(b[k].yy, b[k].x));
      }
    for (int j = i + 1; j <= n; ++j) {
      int l = a[j].x, r = a[j].xx;
      while (!pq.empty() && pq.top().y < a[j].y) {
        add(pq.top().pos, -1);
        pq.pop();
      }
      int res = get(r) - get(l - 1);
      ans += 1ll * res * (res - 1) / 2;
    }
  }
  cout << ans << endl;
}

### 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;
int T;
struct Line {
  int x, y, xx, yy;
  Line(int X = 0, int Y = 0, int XX = 0, int YY = 0)
      : x(X), y(Y), xx(XX), yy(YY) {}
};
int n, m;
Line a[5005], b[5005];
bool cmp1(Line A, Line B) { return A.y < B.y; }
bool cmp2(Line A, Line B) { return A.x < B.x; }
int c[10005];
void add(int x, int v) {
  for (; x <= 10003; x += x & (-x)) c[x] += v;
}
int get(int x) {
  int ans = 0;
  for (; x; x -= x & (-x)) ans += c[x];
  return ans;
}
struct node {
  int y, pos;
  node(int Y = 0, int POS = 0) : y(Y), pos(POS) {}
};
bool operator<(node A, node B) { return A.y > B.y; }
int main() {
  scanf("%d", &T);
  for (int i = 1; i <= T; ++i) {
    int x, y, xx, yy;
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    if (x > xx) swap(x, xx);
    if (y > yy) swap(y, yy);
    x += 5002, y += 5002, xx += 5002, yy += 5002;
    if (y == yy)
      a[++n] = Line(x, y, xx, yy);
    else
      b[++m] = Line(x, y, xx, yy);
  }
  sort(a + 1, a + n + 1, cmp1);
  sort(b + 1, b + m + 1, cmp2);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    priority_queue<node> pq;
    memset(c, 0, sizeof(c));
    for (int k = 1; k <= m; ++k)
      if (a[i].x <= b[k].x && b[k].x <= a[i].xx && b[k].y <= a[i].y &&
          a[i].y <= b[k].yy) {
        add(b[k].x, 1);
        pq.push(node(b[k].yy, b[k].x));
      }
    for (int j = i + 1; j <= n; ++j) {
      int l = a[j].x, r = a[j].xx;
      while (!pq.empty() && pq.top().y < a[j].y) {
        add(pq.top().pos, -1);
        pq.pop();
      }
      int res = get(r) - get(l - 1);
      ans += 1ll * res * (res - 1) / 2;
    }
  }
  cout << ans << endl;
}

```