#include <bits/stdc++.h>
using namespace std;
const long long N = 2e4 + 10, mod = 1e9 + 7;
struct sc {
  int x1, y1, x2, y2;
};
vector<sc> ve[2];
bitset<5010> bt[N], se;
bool ok(int i, int j) {
  if (ve[0][i].x1 < ve[1][j].x1 || ve[0][i].x1 > ve[1][j].x2) return 0;
  if (ve[1][j].y1 < ve[0][i].y1 || ve[1][j].y1 > ve[0][i].y2) return 0;
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    if (a > c) swap(c, a), swap(b, d);
    if (b > d) swap(c, a), swap(b, d);
    if (a == c)
      ve[0].push_back({a, b, c, d});
    else
      ve[1].push_back({a, b, c, d});
  }
  if (ve[0].size() > ve[1].size()) {
    swap(ve[0], ve[1]);
    for (auto &p : ve[0]) {
      swap(p.x1, p.y1);
      swap(p.x2, p.y2);
    }
    for (auto &p : ve[1]) {
      swap(p.x1, p.y1);
      swap(p.x2, p.y2);
    }
  }
  for (int i = 0; i < ve[0].size(); i++) {
    for (int j = 0; j < ve[1].size(); j++) {
      if (ok(i, j)) {
        bt[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < ve[0].size(); i++) {
    for (int j = i + 1; j < ve[0].size(); j++) {
      se = bt[i] & bt[j];
      long long cnt = se.count();
      ans += (cnt - 1) * cnt / 2;
    }
  }
  printf("%lld\n", ans);
}

### Prompt

Create a solution in cpp for 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 long long N = 2e4 + 10, mod = 1e9 + 7;
struct sc {
  int x1, y1, x2, y2;
};
vector<sc> ve[2];
bitset<5010> bt[N], se;
bool ok(int i, int j) {
  if (ve[0][i].x1 < ve[1][j].x1 || ve[0][i].x1 > ve[1][j].x2) return 0;
  if (ve[1][j].y1 < ve[0][i].y1 || ve[1][j].y1 > ve[0][i].y2) return 0;
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    if (a > c) swap(c, a), swap(b, d);
    if (b > d) swap(c, a), swap(b, d);
    if (a == c)
      ve[0].push_back({a, b, c, d});
    else
      ve[1].push_back({a, b, c, d});
  }
  if (ve[0].size() > ve[1].size()) {
    swap(ve[0], ve[1]);
    for (auto &p : ve[0]) {
      swap(p.x1, p.y1);
      swap(p.x2, p.y2);
    }
    for (auto &p : ve[1]) {
      swap(p.x1, p.y1);
      swap(p.x2, p.y2);
    }
  }
  for (int i = 0; i < ve[0].size(); i++) {
    for (int j = 0; j < ve[1].size(); j++) {
      if (ok(i, j)) {
        bt[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < ve[0].size(); i++) {
    for (int j = i + 1; j < ve[0].size(); j++) {
      se = bt[i] & bt[j];
      long long cnt = se.count();
      ans += (cnt - 1) * cnt / 2;
    }
  }
  printf("%lld\n", ans);
}

```