#include <bits/stdc++.h>
const int MX = 1e4 + 23;
int Pcnt;
struct POINT {
  int type, x, y, down;
  bool operator<(const POINT& B) const { return y < B.y; }
} P[MX];
int Hcnt;
struct Hseg {
  int l, r, y;
  bool operator<(const Hseg& B) const { return y < B.y; }
} H[MX];
struct BIT {
  int data[MX];
  void clear() { memset(data, 0, sizeof data); }
  void add(int x, int v) {
    while (x < MX) data[x] += v, x += x & -x;
  }
  int sum(int x) {
    int s = 0;
    while (x > 0) s += data[x], x -= x & -x;
    return s;
  }
} C;
int main() {
  int n;
  std::cin >> n;
  for (int i = 1, x1, y1, x2, y2; i <= n; ++i) {
    std::cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001, x2 += 5001;
    if (x1 > x2) std::swap(x1, x2);
    if (y1 > y2) std::swap(y1, y2);
    if (x1 == x2) {
      P[++Pcnt] = (POINT){1, x1, y1, y1};
      P[++Pcnt] = (POINT){-1, x1, y2 + 1, y1};
    } else {
      H[++Hcnt] = (Hseg){x1, x2, y1};
    }
  }
  long long Ans = 0;
  std::sort(H + 1, H + 1 + Hcnt);
  std::sort(P + 1, P + 1 + Pcnt);
  for (int i = 1; i <= Hcnt; ++i) {
    C.clear();
    int k = 1;
    for (int j = i + 1; j <= Hcnt; ++j) {
      if (H[j].l > H[i].r || H[j].r < H[i].l) continue;
      while (k <= Pcnt && P[k].y <= H[j].y) {
        if (H[i].l <= P[k].x && P[k].x <= H[i].r && P[k].down <= H[i].y) {
          C.add(P[k].x, P[k].type);
        }
        ++k;
      }
      int l = std::max(H[i].l, H[j].l);
      int r = std::min(H[i].r, H[j].r);
      int s = C.sum(r) - C.sum(l - 1);
      Ans += s * (s - 1) / 2;
    }
  }
  std::cout << Ans << std::endl;
  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>
const int MX = 1e4 + 23;
int Pcnt;
struct POINT {
  int type, x, y, down;
  bool operator<(const POINT& B) const { return y < B.y; }
} P[MX];
int Hcnt;
struct Hseg {
  int l, r, y;
  bool operator<(const Hseg& B) const { return y < B.y; }
} H[MX];
struct BIT {
  int data[MX];
  void clear() { memset(data, 0, sizeof data); }
  void add(int x, int v) {
    while (x < MX) data[x] += v, x += x & -x;
  }
  int sum(int x) {
    int s = 0;
    while (x > 0) s += data[x], x -= x & -x;
    return s;
  }
} C;
int main() {
  int n;
  std::cin >> n;
  for (int i = 1, x1, y1, x2, y2; i <= n; ++i) {
    std::cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001, x2 += 5001;
    if (x1 > x2) std::swap(x1, x2);
    if (y1 > y2) std::swap(y1, y2);
    if (x1 == x2) {
      P[++Pcnt] = (POINT){1, x1, y1, y1};
      P[++Pcnt] = (POINT){-1, x1, y2 + 1, y1};
    } else {
      H[++Hcnt] = (Hseg){x1, x2, y1};
    }
  }
  long long Ans = 0;
  std::sort(H + 1, H + 1 + Hcnt);
  std::sort(P + 1, P + 1 + Pcnt);
  for (int i = 1; i <= Hcnt; ++i) {
    C.clear();
    int k = 1;
    for (int j = i + 1; j <= Hcnt; ++j) {
      if (H[j].l > H[i].r || H[j].r < H[i].l) continue;
      while (k <= Pcnt && P[k].y <= H[j].y) {
        if (H[i].l <= P[k].x && P[k].x <= H[i].r && P[k].down <= H[i].y) {
          C.add(P[k].x, P[k].type);
        }
        ++k;
      }
      int l = std::max(H[i].l, H[j].l);
      int r = std::min(H[i].r, H[j].r);
      int s = C.sum(r) - C.sum(l - 1);
      Ans += s * (s - 1) / 2;
    }
  }
  std::cout << Ans << std::endl;
  return 0;
}

```