#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
const int mo = 1e9 + 7;
struct QS {
  int x, yl, yr;
  friend bool operator<(const QS& a, const QS& b) { return a.x < b.x; }
};
struct FC {
  int x, y, id;
  friend bool operator<(const FC& a, const FC& b) { return a.x < b.x; }
};
int a[N];
bool v[N];
inline int q(int x) {
  int r = 0;
  for (; x; x -= x & (-x)) r += a[x];
  return r;
}
inline void f(int x, int y) {
  for (; x < N; x += x & (-x)) a[x] += y;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<QS> qs;
  vector<FC> fc1, fc2;
  for (auto i = (1); i <= (n); ++i) {
    int xl, xr, yl, yr;
    scanf("%d%d%d%d", &xl, &yl, &xr, &yr);
    xl += 5001;
    xr += 5001;
    yl += 5001;
    yr += 5001;
    if (xl == xr) {
      if (yl > yr) swap(yl, yr);
      qs.push_back({xl, yl, yr});
    } else {
      if (xl > xr) swap(xl, xr);
      fc1.push_back({xl, yl, i});
      fc2.push_back({xr, yl, i});
    }
  }
  long long ans = 0;
  sort(qs.begin(), qs.end());
  sort(fc1.begin(), fc1.end());
  sort(fc2.begin(), fc2.end());
  auto x = fc1.begin(), y = fc2.begin();
  for (auto i = qs.begin(); i != qs.end(); ++i) {
    for (; x != fc1.end() && x->x <= i->x; ++x) {
      v[x->id] = 1;
      f(x->y, 1);
    }
    for (; y != fc2.end() && y->x < i->x; ++y) {
      v[y->id] = 0;
      f(y->y, -1);
    }
    auto z = y;
    for (auto j = i + 1; j != qs.end(); ++j) {
      int l = max(i->yl, j->yl), r = min(i->yr, j->yr);
      if (l > r) continue;
      for (; z != fc2.end() && z->x < j->x; ++z)
        if (v[z->id]) f(z->y, -1);
      long long s = q(r) - q(l - 1);
      ans += s * (s - 1) / 2;
    }
    while (z-- != y)
      if (v[z->id]) f(z->y, 1);
  }
  printf("%lld\n", ans);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 = 1e4 + 10;
const int mo = 1e9 + 7;
struct QS {
  int x, yl, yr;
  friend bool operator<(const QS& a, const QS& b) { return a.x < b.x; }
};
struct FC {
  int x, y, id;
  friend bool operator<(const FC& a, const FC& b) { return a.x < b.x; }
};
int a[N];
bool v[N];
inline int q(int x) {
  int r = 0;
  for (; x; x -= x & (-x)) r += a[x];
  return r;
}
inline void f(int x, int y) {
  for (; x < N; x += x & (-x)) a[x] += y;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<QS> qs;
  vector<FC> fc1, fc2;
  for (auto i = (1); i <= (n); ++i) {
    int xl, xr, yl, yr;
    scanf("%d%d%d%d", &xl, &yl, &xr, &yr);
    xl += 5001;
    xr += 5001;
    yl += 5001;
    yr += 5001;
    if (xl == xr) {
      if (yl > yr) swap(yl, yr);
      qs.push_back({xl, yl, yr});
    } else {
      if (xl > xr) swap(xl, xr);
      fc1.push_back({xl, yl, i});
      fc2.push_back({xr, yl, i});
    }
  }
  long long ans = 0;
  sort(qs.begin(), qs.end());
  sort(fc1.begin(), fc1.end());
  sort(fc2.begin(), fc2.end());
  auto x = fc1.begin(), y = fc2.begin();
  for (auto i = qs.begin(); i != qs.end(); ++i) {
    for (; x != fc1.end() && x->x <= i->x; ++x) {
      v[x->id] = 1;
      f(x->y, 1);
    }
    for (; y != fc2.end() && y->x < i->x; ++y) {
      v[y->id] = 0;
      f(y->y, -1);
    }
    auto z = y;
    for (auto j = i + 1; j != qs.end(); ++j) {
      int l = max(i->yl, j->yl), r = min(i->yr, j->yr);
      if (l > r) continue;
      for (; z != fc2.end() && z->x < j->x; ++z)
        if (v[z->id]) f(z->y, -1);
      long long s = q(r) - q(l - 1);
      ans += s * (s - 1) / 2;
    }
    while (z-- != y)
      if (v[z->id]) f(z->y, 1);
  }
  printf("%lld\n", ans);
}

```