#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % MOD;
}
inline long long Mum(initializer_list<long long> ls) {
  long long a = 1;
  for (long long b : ls) a = a * b % MOD;
  return a;
}
const int N = 5e4 + 100;
struct bit {
  long long c[N * 2];
  void Clear() { memset(c, 0, sizeof(c)); }
  int lowbit(int x) { return x & -x; }
  long long Ask(int x) {
    long long res = 0;
    while (x) res += c[x], x -= lowbit(x);
    return res;
  }
  long long Ask(int l, int r) {
    l += 5100, r += 5100;
    return Ask(r) - Ask(l - 1);
  }
  void Add(int x, long long v) {
    x += 5100;
    while (x < N) c[x] += v, x += lowbit(x);
  }
} tre;
struct node {
  int a, l, r;
  bool operator<(const node &o) const { return a < o.a; }
};
vector<node> h, l;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      l.push_back({a, b, d});
    else
      h.push_back({b, a, c});
  }
  sort((h).begin(), (h).end());
  sort((l).begin(), (l).end(),
       [](const node &a, const node &b) { return a.r < b.r; });
  long long ans = 0;
  for (int i = 0; i < ((int)(h).size()); ++i) {
    int y = h[i].a, L = h[i].l, R = h[i].r;
    vector<node> v = l;
    tre.Clear();
    for (int j = ((int)(h).size()) - 1; j > i; --j) {
      while (!v.empty() && v.back().r >= h[j].a) {
        if (v.back().l <= y && v.back().a >= L && v.back().a <= R)
          tre.Add(v.back().a, 1);
        v.pop_back();
      }
      long long res = tre.Ask(h[j].l, h[j].r);
      ans += res * (res - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = 1LL * a * b % MOD;
}
inline long long Mum(initializer_list<long long> ls) {
  long long a = 1;
  for (long long b : ls) a = a * b % MOD;
  return a;
}
const int N = 5e4 + 100;
struct bit {
  long long c[N * 2];
  void Clear() { memset(c, 0, sizeof(c)); }
  int lowbit(int x) { return x & -x; }
  long long Ask(int x) {
    long long res = 0;
    while (x) res += c[x], x -= lowbit(x);
    return res;
  }
  long long Ask(int l, int r) {
    l += 5100, r += 5100;
    return Ask(r) - Ask(l - 1);
  }
  void Add(int x, long long v) {
    x += 5100;
    while (x < N) c[x] += v, x += lowbit(x);
  }
} tre;
struct node {
  int a, l, r;
  bool operator<(const node &o) const { return a < o.a; }
};
vector<node> h, l;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c)
      l.push_back({a, b, d});
    else
      h.push_back({b, a, c});
  }
  sort((h).begin(), (h).end());
  sort((l).begin(), (l).end(),
       [](const node &a, const node &b) { return a.r < b.r; });
  long long ans = 0;
  for (int i = 0; i < ((int)(h).size()); ++i) {
    int y = h[i].a, L = h[i].l, R = h[i].r;
    vector<node> v = l;
    tre.Clear();
    for (int j = ((int)(h).size()) - 1; j > i; --j) {
      while (!v.empty() && v.back().r >= h[j].a) {
        if (v.back().l <= y && v.back().a >= L && v.back().a <= R)
          tre.Add(v.back().a, 1);
        v.pop_back();
      }
      long long res = tre.Ask(h[j].l, h[j].r);
      ans += res * (res - 1) / 2;
    }
  }
  cout << ans << endl;
  return 0;
}

```