#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  short n, a, b, c, d;
  cin >> n;
  vector<pair<short, pair<short, short>>> x, y;
  for (short i = 0; i < n; i += 1) {
    cin >> a >> b >> c >> d;
    if (a > c) {
      swap(a, c);
    }
    if (b > d) {
      swap(b, d);
    }
    if (a == c) {
      x.push_back({c, {b, d}});
    } else {
      y.push_back({b, {a, c}});
    }
  }
  if (y.size() > 100) {
    int siz = x.size();
    bitset<5005> z[siz];
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = 0; j < y.size(); j += 1) {
        if (x[i].second.first <= y[j].first &&
            x[i].second.second >= y[j].first) {
          if (y[j].second.first <= x[i].first &&
              y[j].second.second >= x[i].first) {
            z[i][j] = 1;
          }
        }
      }
    }
    long long ans = 0;
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = i + 1; j < x.size(); j += 1) {
        long long val = (z[i] & z[j]).count();
        ans += val * (val - 1) / 2;
      }
    }
    cout << ans;
  } else {
    int siz = x.size();
    bitset<105> z[siz];
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = 0; j < y.size(); j += 1) {
        if (x[i].second.first <= y[j].first &&
            x[i].second.second >= y[j].first) {
          if (y[j].second.first <= x[i].first &&
              y[j].second.second >= x[i].first) {
            z[i][j] = 1;
          }
        }
      }
    }
    long long ans = 0;
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = i + 1; j < x.size(); j += 1) {
        long long val = (z[i] & z[j]).count();
        ans += val * (val - 1) / 2;
      }
    }
    cout << ans;
  }
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  short n, a, b, c, d;
  cin >> n;
  vector<pair<short, pair<short, short>>> x, y;
  for (short i = 0; i < n; i += 1) {
    cin >> a >> b >> c >> d;
    if (a > c) {
      swap(a, c);
    }
    if (b > d) {
      swap(b, d);
    }
    if (a == c) {
      x.push_back({c, {b, d}});
    } else {
      y.push_back({b, {a, c}});
    }
  }
  if (y.size() > 100) {
    int siz = x.size();
    bitset<5005> z[siz];
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = 0; j < y.size(); j += 1) {
        if (x[i].second.first <= y[j].first &&
            x[i].second.second >= y[j].first) {
          if (y[j].second.first <= x[i].first &&
              y[j].second.second >= x[i].first) {
            z[i][j] = 1;
          }
        }
      }
    }
    long long ans = 0;
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = i + 1; j < x.size(); j += 1) {
        long long val = (z[i] & z[j]).count();
        ans += val * (val - 1) / 2;
      }
    }
    cout << ans;
  } else {
    int siz = x.size();
    bitset<105> z[siz];
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = 0; j < y.size(); j += 1) {
        if (x[i].second.first <= y[j].first &&
            x[i].second.second >= y[j].first) {
          if (y[j].second.first <= x[i].first &&
              y[j].second.second >= x[i].first) {
            z[i][j] = 1;
          }
        }
      }
    }
    long long ans = 0;
    for (short i = 0; i < x.size(); i += 1) {
      for (short j = i + 1; j < x.size(); j += 1) {
        long long val = (z[i] & z[j]).count();
        ans += val * (val - 1) / 2;
      }
    }
    cout << ans;
  }
}

```