#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
void ECHO(string _s) {
  cout << endl;
  (void)_s;
}
template <typename T, typename... Args>
void ECHO(string _s, T x, Args... args) {
  int _i;
  string _s2 = "";
  for (_i = 0; _i < (int)(_s).size(); ++_i) {
    if (_s[_i] == ',') break;
    if (_s[_i] != ' ') _s2 += _s[_i];
  }
  if (_i == (int)(_s).size()) {
    --_i;
  }
  cout << "  (" << _s2 << "):" << x;
  ECHO(_s.substr(_i + 1, (int)(_s).size() - _i - 1), args...);
}
template <typename T0, typename T1>
inline ostream& operator<<(ostream& os, pair<T0, T1>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  for (int i = 0; i < (int)(v).size(); ++i) {
    if (i > 0) cout << "_";
    cout << v[i];
  }
  cout << endl;
  return os;
}
inline long long _gcd(long long a, long long b) {
  while (b) b %= a ^= b ^= a ^= b;
  return a;
}
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long inf = LLONG_MAX / 10;
const int int_inf = INT_MAX / 10;
bool intersect(
    pair<pair<long long, long long>, pair<long long, long long>>& a,
    pair<pair<long long, long long>, pair<long long, long long>>& b) {
  if (a.first.first <= b.first.first && b.first.first <= a.second.first &&
      b.first.second <= a.first.second && a.first.second <= b.second.second)
    return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, n, a, b, c, d;
  cin >> n;
  vector<pair<pair<long long, long long>, pair<long long, long long>>> v, h;
  for (i = 0; i < n; i++) {
    cin >> a >> b >> c >> d;
    if (a == c) {
      if (b > d)
        v.push_back(make_pair(make_pair(c, d), make_pair(a, b)));
      else
        v.push_back(make_pair(make_pair(a, b), make_pair(c, d)));
    } else {
      if (a > c)
        h.push_back(make_pair(make_pair(c, d), make_pair(a, b)));
      else
        h.push_back(make_pair(make_pair(a, b), make_pair(c, d)));
    }
  }
  const long long hs = h.size();
  bitset<5005> ma[hs];
  long long cont = 0, num;
  for (i = 0; i < (int)v.size(); i++) {
    for (j = 0; j < (int)h.size(); j++) {
      if (intersect(h[j], v[i]) == true) {
        ma[j][i].flip();
      }
    }
  }
  cont = 0;
  for (i = 0; i < hs; i++) {
    for (j = i + 1; j < hs; j++) {
      num = (ma[i] & ma[j]).count();
      cont += (num * (num - 1)) / 2;
    }
  }
  cout << cont << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
void ECHO(string _s) {
  cout << endl;
  (void)_s;
}
template <typename T, typename... Args>
void ECHO(string _s, T x, Args... args) {
  int _i;
  string _s2 = "";
  for (_i = 0; _i < (int)(_s).size(); ++_i) {
    if (_s[_i] == ',') break;
    if (_s[_i] != ' ') _s2 += _s[_i];
  }
  if (_i == (int)(_s).size()) {
    --_i;
  }
  cout << "  (" << _s2 << "):" << x;
  ECHO(_s.substr(_i + 1, (int)(_s).size() - _i - 1), args...);
}
template <typename T0, typename T1>
inline ostream& operator<<(ostream& os, pair<T0, T1>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  for (int i = 0; i < (int)(v).size(); ++i) {
    if (i > 0) cout << "_";
    cout << v[i];
  }
  cout << endl;
  return os;
}
inline long long _gcd(long long a, long long b) {
  while (b) b %= a ^= b ^= a ^= b;
  return a;
}
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long inf = LLONG_MAX / 10;
const int int_inf = INT_MAX / 10;
bool intersect(
    pair<pair<long long, long long>, pair<long long, long long>>& a,
    pair<pair<long long, long long>, pair<long long, long long>>& b) {
  if (a.first.first <= b.first.first && b.first.first <= a.second.first &&
      b.first.second <= a.first.second && a.first.second <= b.second.second)
    return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, n, a, b, c, d;
  cin >> n;
  vector<pair<pair<long long, long long>, pair<long long, long long>>> v, h;
  for (i = 0; i < n; i++) {
    cin >> a >> b >> c >> d;
    if (a == c) {
      if (b > d)
        v.push_back(make_pair(make_pair(c, d), make_pair(a, b)));
      else
        v.push_back(make_pair(make_pair(a, b), make_pair(c, d)));
    } else {
      if (a > c)
        h.push_back(make_pair(make_pair(c, d), make_pair(a, b)));
      else
        h.push_back(make_pair(make_pair(a, b), make_pair(c, d)));
    }
  }
  const long long hs = h.size();
  bitset<5005> ma[hs];
  long long cont = 0, num;
  for (i = 0; i < (int)v.size(); i++) {
    for (j = 0; j < (int)h.size(); j++) {
      if (intersect(h[j], v[i]) == true) {
        ma[j][i].flip();
      }
    }
  }
  cont = 0;
  for (i = 0; i < hs; i++) {
    for (j = i + 1; j < hs; j++) {
      num = (ma[i] & ma[j]).count();
      cont += (num * (num - 1)) / 2;
    }
  }
  cout << cont << "\n";
  return 0;
}

```