#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first - a.second) > (b.first - b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<pair<pair<int, int>, pair<int, int>>> hor, ver;
  for (int i = 0; i < n; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    pair<int, int> t1 = {a, b};
    pair<int, int> t2 = {c, d};
    if (b == d)
      hor.push_back({min(t1, t2), max(t1, t2)});
    else
      ver.push_back({min(t1, t2), max(t1, t2)});
    ;
  }
  sort(hor.begin(), hor.end());
  sort(ver.begin(), ver.end());
  int vern = ver.size();
  vector<bitset<5000>> btst(vern);
  for (int i = 0; i < ver.size(); ++i) {
    for (int j = 0; j < hor.size(); j++) {
      int lev = hor[j].first.second;
      int a = hor[j].first.first;
      int b = hor[j].second.first;
      int pl = ver[i].first.first;
      int p = ver[i].first.second;
      int q = ver[i].second.second;
      bool flag = true;
      if (!(lev >= min(p, q) && lev <= max(p, q))) flag = false;
      if (!(pl >= min(a, b) && pl <= max(a, b))) flag = false;
      if (flag) btst[i][j] = 1;
    }
  }
  long long int ans = 0;
  for (int i = 0; i < ver.size(); ++i) {
    for (int j = i + 1; j < ver.size(); ++j) {
      int cn = (btst[i] & btst[j]).count();
      ans += (cn * (cn - 1)) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first - a.second) > (b.first - b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<pair<pair<int, int>, pair<int, int>>> hor, ver;
  for (int i = 0; i < n; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    pair<int, int> t1 = {a, b};
    pair<int, int> t2 = {c, d};
    if (b == d)
      hor.push_back({min(t1, t2), max(t1, t2)});
    else
      ver.push_back({min(t1, t2), max(t1, t2)});
    ;
  }
  sort(hor.begin(), hor.end());
  sort(ver.begin(), ver.end());
  int vern = ver.size();
  vector<bitset<5000>> btst(vern);
  for (int i = 0; i < ver.size(); ++i) {
    for (int j = 0; j < hor.size(); j++) {
      int lev = hor[j].first.second;
      int a = hor[j].first.first;
      int b = hor[j].second.first;
      int pl = ver[i].first.first;
      int p = ver[i].first.second;
      int q = ver[i].second.second;
      bool flag = true;
      if (!(lev >= min(p, q) && lev <= max(p, q))) flag = false;
      if (!(pl >= min(a, b) && pl <= max(a, b))) flag = false;
      if (flag) btst[i][j] = 1;
    }
  }
  long long int ans = 0;
  for (int i = 0; i < ver.size(); ++i) {
    for (int j = i + 1; j < ver.size(); ++j) {
      int cn = (btst[i] & btst[j]).count();
      ans += (cn * (cn - 1)) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

```