#include <bits/stdc++.h>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(322179);
}
const int C = 5010;
int tr[C * 8];
bool cross(pair<pair<int, int>, int> hor, pair<int, pair<int, int> > vert) {
  return (vert.second.first <= hor.second &&
          hor.second <= vert.second.second) &&
         (hor.first.first <= vert.first && vert.first <= hor.first.second);
}
void init() {
  for (int i = 0; i < C * 8; ++i) {
    tr[i] = 0;
  }
}
int get(int l, int r, int lb = -C, int rb = C, int v = 0) {
  if (l == lb && r == rb) {
    return tr[v];
  } else {
    int mb = (lb + rb) / 2;
    if (r <= mb) {
      return get(l, r, lb, mb, v * 2 + 1);
    } else if (l >= mb) {
      return get(l, r, mb, rb, v * 2 + 2);
    } else {
      return get(l, mb, lb, mb, v * 2 + 1) + get(mb, r, mb, rb, v * 2 + 2);
    }
  }
}
void add(int i, int x, int lb = -C, int rb = C, int v = 0) {
  if (lb + 1 == rb) {
    tr[v] += x;
  } else {
    int mb = (lb + rb) / 2;
    if (i < mb) {
      add(i, x, lb, mb, v * 2 + 1);
    } else if (i >= mb) {
      add(i, x, mb, rb, v * 2 + 2);
    }
    tr[v] = tr[v * 2 + 1] + tr[v * 2 + 2];
  }
}
signed main() {
  doRoutine();
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > verts;
  vector<pair<pair<int, int>, int> > hors;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) {
      swap(x1, x2);
    }
    if (y1 > y2) {
      swap(y1, y2);
    }
    if (x1 == x2) {
      verts.push_back({x1, {y1, y2}});
    } else {
      hors.push_back({{x1, x2}, y1});
    }
  }
  sort(verts.begin(), verts.end());
  long long ans = 0;
  for (int i = 0; i < verts.size(); ++i) {
    vector<pair<int, int> > events;
    for (auto seg : hors) {
      if (cross(seg, verts[i])) {
        events.push_back({seg.first.second, seg.second});
      }
    }
    if (!events.empty()) {
      sort(events.begin(), events.end());
      reverse(events.begin(), events.end());
      int ptr = 0;
      init();
      for (int j = (int)verts.size() - 1; j > i; --j) {
        while ((ptr < events.size() && events[ptr].first >= verts[j].first)) {
          add(events[ptr].second, 1);
          ++ptr;
        }
        int cnt = get(verts[j].second.first, verts[j].second.second + 1);
        ans += cnt * (cnt - 1) / 2;
      }
    }
  }
  cout << ans << 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>
using namespace std;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(322179);
}
const int C = 5010;
int tr[C * 8];
bool cross(pair<pair<int, int>, int> hor, pair<int, pair<int, int> > vert) {
  return (vert.second.first <= hor.second &&
          hor.second <= vert.second.second) &&
         (hor.first.first <= vert.first && vert.first <= hor.first.second);
}
void init() {
  for (int i = 0; i < C * 8; ++i) {
    tr[i] = 0;
  }
}
int get(int l, int r, int lb = -C, int rb = C, int v = 0) {
  if (l == lb && r == rb) {
    return tr[v];
  } else {
    int mb = (lb + rb) / 2;
    if (r <= mb) {
      return get(l, r, lb, mb, v * 2 + 1);
    } else if (l >= mb) {
      return get(l, r, mb, rb, v * 2 + 2);
    } else {
      return get(l, mb, lb, mb, v * 2 + 1) + get(mb, r, mb, rb, v * 2 + 2);
    }
  }
}
void add(int i, int x, int lb = -C, int rb = C, int v = 0) {
  if (lb + 1 == rb) {
    tr[v] += x;
  } else {
    int mb = (lb + rb) / 2;
    if (i < mb) {
      add(i, x, lb, mb, v * 2 + 1);
    } else if (i >= mb) {
      add(i, x, mb, rb, v * 2 + 2);
    }
    tr[v] = tr[v * 2 + 1] + tr[v * 2 + 2];
  }
}
signed main() {
  doRoutine();
  int n;
  cin >> n;
  vector<pair<int, pair<int, int> > > verts;
  vector<pair<pair<int, int>, int> > hors;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 > x2) {
      swap(x1, x2);
    }
    if (y1 > y2) {
      swap(y1, y2);
    }
    if (x1 == x2) {
      verts.push_back({x1, {y1, y2}});
    } else {
      hors.push_back({{x1, x2}, y1});
    }
  }
  sort(verts.begin(), verts.end());
  long long ans = 0;
  for (int i = 0; i < verts.size(); ++i) {
    vector<pair<int, int> > events;
    for (auto seg : hors) {
      if (cross(seg, verts[i])) {
        events.push_back({seg.first.second, seg.second});
      }
    }
    if (!events.empty()) {
      sort(events.begin(), events.end());
      reverse(events.begin(), events.end());
      int ptr = 0;
      init();
      for (int j = (int)verts.size() - 1; j > i; --j) {
        while ((ptr < events.size() && events[ptr].first >= verts[j].first)) {
          add(events[ptr].second, 1);
          ++ptr;
        }
        int cnt = get(verts[j].second.first, verts[j].second.second + 1);
        ans += cnt * (cnt - 1) / 2;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```