#include <bits/stdc++.h>
using namespace std;
struct seg {
  long long x1, y1, x2, y2;
};
bool cmp(const seg &i, const seg &j) { return i.x1 < j.x1; }
struct fenwickTree {
  vector<long long> bit;
  long long n;
  fenwickTree(long long _n) {
    n = _n;
    bit.assign(n, 0);
  }
  long long sum(long long r) {
    long long ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
    return ret;
  }
  void add(long long idx, long long d) {
    for (; idx < n; idx = idx | (idx + 1)) bit[idx] += d;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<seg> ver, hor;
  for (long long i = 0; i < n; ++i) {
    long long x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      ver.push_back({x1, y1, x2, y2});
    } else {
      hor.push_back({x1, y1, x2, y2});
    }
  }
  sort(ver.begin(), ver.end(), cmp);
  sort(hor.begin(), hor.end(), cmp);
  long long res = 0;
  for (long long i = 0; i < (long long)ver.size(); ++i) {
    long long posHor = 0;
    fenwickTree ft(10010);
    for (long long j = 0; j < i; ++j) {
      while (posHor < (long long)hor.size() && hor[posHor].x1 <= ver[j].x1) {
        if (ver[i].y1 <= hor[posHor].y1 && hor[posHor].y2 <= ver[i].y2) {
          if (hor[posHor].x2 >= ver[i].x2) {
            ft.add(hor[posHor].y1 + 5000, 1);
          }
        }
        ++posHor;
      }
      long long L = max(ver[i].y1, ver[j].y1) + 5000;
      long long R = min(ver[i].y2, ver[j].y2) + 5000;
      if (L <= R) {
        long long cnt = ft.sum(L, R);
        res += cnt * (cnt - 1) / 2;
      }
    }
  }
  cout << res << '\n';
}

### Prompt

Your challenge is to write 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;
struct seg {
  long long x1, y1, x2, y2;
};
bool cmp(const seg &i, const seg &j) { return i.x1 < j.x1; }
struct fenwickTree {
  vector<long long> bit;
  long long n;
  fenwickTree(long long _n) {
    n = _n;
    bit.assign(n, 0);
  }
  long long sum(long long r) {
    long long ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
    return ret;
  }
  void add(long long idx, long long d) {
    for (; idx < n; idx = idx | (idx + 1)) bit[idx] += d;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
};
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<seg> ver, hor;
  for (long long i = 0; i < n; ++i) {
    long long x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      ver.push_back({x1, y1, x2, y2});
    } else {
      hor.push_back({x1, y1, x2, y2});
    }
  }
  sort(ver.begin(), ver.end(), cmp);
  sort(hor.begin(), hor.end(), cmp);
  long long res = 0;
  for (long long i = 0; i < (long long)ver.size(); ++i) {
    long long posHor = 0;
    fenwickTree ft(10010);
    for (long long j = 0; j < i; ++j) {
      while (posHor < (long long)hor.size() && hor[posHor].x1 <= ver[j].x1) {
        if (ver[i].y1 <= hor[posHor].y1 && hor[posHor].y2 <= ver[i].y2) {
          if (hor[posHor].x2 >= ver[i].x2) {
            ft.add(hor[posHor].y1 + 5000, 1);
          }
        }
        ++posHor;
      }
      long long L = max(ver[i].y1, ver[j].y1) + 5000;
      long long R = min(ver[i].y2, ver[j].y2) + 5000;
      if (L <= R) {
        long long cnt = ft.sum(L, R);
        res += cnt * (cnt - 1) / 2;
      }
    }
  }
  cout << res << '\n';
}

```