#include <bits/stdc++.h>
using namespace std;
const int N = 5000 + 7;
const int C = 2 * N;
int aib[C];
void add(int p, int x) {
  p += N;
  for (int i = p; i < C; i += i & (-i)) aib[i] += x;
}
int prefix(int p) {
  p += N;
  int r = 0;
  for (int i = p; i > 0; i -= i & (-i)) r += aib[i];
  return r;
}
struct Point {
  int x;
  int y;
  int i;
};
bool operator<(Point a, Point b) { return a.y < b.y; }
int cnt_vert;
Point in[N], out[N];
struct Segment {
  int y;
  int x1, x2;
};
bool operator<(Segment a, Segment b) { return a.y < b.y; }
int cnt_orz;
Segment orz[N];
bool used[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int all;
  cin >> all;
  for (int i = 1; i <= all; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      cnt_vert++;
      in[cnt_vert] = {x1, min(y1, y2), cnt_vert};
      out[cnt_vert] = {x2, max(y1, y2), cnt_vert};
    } else
      orz[++cnt_orz] = {y1, min(x1, x2), max(x1, x2)};
  }
  sort(in + 1, in + cnt_vert + 1);
  sort(out + 1, out + cnt_vert + 1);
  sort(orz + 1, orz + cnt_orz + 1);
  long long sol = 0;
  for (int i = 1; i <= cnt_orz; i++) {
    for (int j = 0; j < C; j++) aib[j] = 0;
    for (int j = 1; j <= cnt_vert; j++)
      if (in[j].y <= orz[i].y)
        used[in[j].i] = 1, add(in[j].x, 1);
      else
        used[in[j].i] = 0;
    int cur_out = 1;
    for (int j = i + 1; j <= cnt_orz; j++) {
      while (cur_out <= cnt_vert && out[cur_out].y < orz[j].y) {
        if (used[out[cur_out].i]) add(out[cur_out].x, -1);
        cur_out++;
      }
      int st = max(orz[i].x1, orz[j].x1);
      int dr = min(orz[i].x2, orz[j].x2);
      if (st <= dr) {
        long long x = prefix(dr) - prefix(st - 1);
        sol += x * (x - 1) / 2;
      }
    }
  }
  cout << sol << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 5000 + 7;
const int C = 2 * N;
int aib[C];
void add(int p, int x) {
  p += N;
  for (int i = p; i < C; i += i & (-i)) aib[i] += x;
}
int prefix(int p) {
  p += N;
  int r = 0;
  for (int i = p; i > 0; i -= i & (-i)) r += aib[i];
  return r;
}
struct Point {
  int x;
  int y;
  int i;
};
bool operator<(Point a, Point b) { return a.y < b.y; }
int cnt_vert;
Point in[N], out[N];
struct Segment {
  int y;
  int x1, x2;
};
bool operator<(Segment a, Segment b) { return a.y < b.y; }
int cnt_orz;
Segment orz[N];
bool used[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int all;
  cin >> all;
  for (int i = 1; i <= all; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2) {
      cnt_vert++;
      in[cnt_vert] = {x1, min(y1, y2), cnt_vert};
      out[cnt_vert] = {x2, max(y1, y2), cnt_vert};
    } else
      orz[++cnt_orz] = {y1, min(x1, x2), max(x1, x2)};
  }
  sort(in + 1, in + cnt_vert + 1);
  sort(out + 1, out + cnt_vert + 1);
  sort(orz + 1, orz + cnt_orz + 1);
  long long sol = 0;
  for (int i = 1; i <= cnt_orz; i++) {
    for (int j = 0; j < C; j++) aib[j] = 0;
    for (int j = 1; j <= cnt_vert; j++)
      if (in[j].y <= orz[i].y)
        used[in[j].i] = 1, add(in[j].x, 1);
      else
        used[in[j].i] = 0;
    int cur_out = 1;
    for (int j = i + 1; j <= cnt_orz; j++) {
      while (cur_out <= cnt_vert && out[cur_out].y < orz[j].y) {
        if (used[out[cur_out].i]) add(out[cur_out].x, -1);
        cur_out++;
      }
      int st = max(orz[i].x1, orz[j].x1);
      int dr = min(orz[i].x2, orz[j].x2);
      if (st <= dr) {
        long long x = prefix(dr) - prefix(st - 1);
        sol += x * (x - 1) / 2;
      }
    }
  }
  cout << sol << "\n";
  return 0;
}

```