#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops,fast-math")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int INT_INF = (int)(2e9);
const ll LL_INF = (ll)(2e18);
const ll mod = 1e9 + 7;
const int NIL = -1;
static mt19937 _g(time(nullptr));
inline ll randint(ll a, ll b) {
  ll w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const pair<T, S> p) {
  cout << "[" << p.first << ";" << p.second << "]";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
bool intr(int l1, int r1, int l2, int r2) {
  if (l1 > r1) swap(l1, r1);
  if (l2 > r2) swap(l2, r2);
  return min(r1, r2) >= max(l1, l2);
}
struct Segment {
  int x1, x2, y1, y2;
  bool intrsect(const Segment& other) const {
    return intr(x1, x2, other.x1, other.x2) && intr(y1, y2, other.y1, other.y2);
  }
};
const int MAX_N = 5007;
vector<Segment> vert, hor;
int n, V, H;
bitset<MAX_N> m[MAX_N];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    Segment s;
    cin >> s.x1 >> s.y1 >> s.x2 >> s.y2;
    if (s.x1 == s.x2)
      vert.push_back(s);
    else
      hor.push_back(s);
  }
  if (((int)(vert).size()) > ((int)(hor).size())) swap(vert, hor);
  V = ((int)(vert).size()), H = ((int)(hor).size());
  for (int i = 0; i < V; ++i)
    for (int j = 0; j < H; ++j) m[i][j] = vert[i].intrsect(hor[j]);
  ll answ = 0;
  for (int i1 = 0; i1 < V; ++i1) {
    for (int i2 = 0; i2 < i1; ++i2) {
      auto a = m[i1] & m[i2];
      ll cnt = a.count();
      answ += (cnt * (cnt - 1)) / 2;
    }
  }
  cout << answ << '\n';
}
int main() {
  fast_io();
  solve();
  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>
#pragma GCC optimize("Ofast,unroll-loops,fast-math")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int INT_INF = (int)(2e9);
const ll LL_INF = (ll)(2e18);
const ll mod = 1e9 + 7;
const int NIL = -1;
static mt19937 _g(time(nullptr));
inline ll randint(ll a, ll b) {
  ll w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const pair<T, S> p) {
  cout << "[" << p.first << ";" << p.second << "]";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
bool intr(int l1, int r1, int l2, int r2) {
  if (l1 > r1) swap(l1, r1);
  if (l2 > r2) swap(l2, r2);
  return min(r1, r2) >= max(l1, l2);
}
struct Segment {
  int x1, x2, y1, y2;
  bool intrsect(const Segment& other) const {
    return intr(x1, x2, other.x1, other.x2) && intr(y1, y2, other.y1, other.y2);
  }
};
const int MAX_N = 5007;
vector<Segment> vert, hor;
int n, V, H;
bitset<MAX_N> m[MAX_N];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    Segment s;
    cin >> s.x1 >> s.y1 >> s.x2 >> s.y2;
    if (s.x1 == s.x2)
      vert.push_back(s);
    else
      hor.push_back(s);
  }
  if (((int)(vert).size()) > ((int)(hor).size())) swap(vert, hor);
  V = ((int)(vert).size()), H = ((int)(hor).size());
  for (int i = 0; i < V; ++i)
    for (int j = 0; j < H; ++j) m[i][j] = vert[i].intrsect(hor[j]);
  ll answ = 0;
  for (int i1 = 0; i1 < V; ++i1) {
    for (int i2 = 0; i2 < i1; ++i2) {
      auto a = m[i1] & m[i2];
      ll cnt = a.count();
      answ += (cnt * (cnt - 1)) / 2;
    }
  }
  cout << answ << '\n';
}
int main() {
  fast_io();
  solve();
  return 0;
}

```