#include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct BitSet {
  vector<unsigned long long> bits;
  BitSet(int n) : bits((n + 63) / 64) {}
  inline void set(int bit) { bits[bit >> 6] |= 1ull << (bit & 63); }
  BitSet operator&(const BitSet &bs) {
    BitSet ans = *this;
    for (int i = 0; i < (int)ans.bits.size(); ++i) ans.bits[i] &= bs.bits[i];
    return ans;
  }
  int count() const {
    int ans = 0;
    for (const auto &v : bits)
      ans += popcount(v >> 32) + popcount(v & (unsigned)-1);
    return ans;
  }

 private:
  static inline int popcount(unsigned x) {
    x = x - ((x >> 1) & 0x55555555);
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
    return ((x + (x >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<tuple<int, int, int>> h, v;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2)
      v.emplace_back(x1, min(y1, y2), max(y1, y2));
    else
      h.emplace_back(y1, min(x1, x2), max(x1, x2));
  }
  int N = h.size(), M = v.size();
  vector<BitSet> bs(N, BitSet(M));
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < M; ++j) {
      int x1, x2, x3;
      int y1, y2, y3;
      tie(y1, x2, x3) = h[i];
      tie(x1, y2, y3) = v[j];
      if (x2 <= x1 && x1 <= x3 && y2 <= y1 && y1 <= y3) bs[i].set(j);
    }
  ll ans = 0;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < i; ++j) {
      int same = (bs[i] & bs[j]).count();
      ans += same * (same - 1) / 2;
    }
  cout << ans << endl;
  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>
using namespace std;
using ll = long long;
struct BitSet {
  vector<unsigned long long> bits;
  BitSet(int n) : bits((n + 63) / 64) {}
  inline void set(int bit) { bits[bit >> 6] |= 1ull << (bit & 63); }
  BitSet operator&(const BitSet &bs) {
    BitSet ans = *this;
    for (int i = 0; i < (int)ans.bits.size(); ++i) ans.bits[i] &= bs.bits[i];
    return ans;
  }
  int count() const {
    int ans = 0;
    for (const auto &v : bits)
      ans += popcount(v >> 32) + popcount(v & (unsigned)-1);
    return ans;
  }

 private:
  static inline int popcount(unsigned x) {
    x = x - ((x >> 1) & 0x55555555);
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
    return ((x + (x >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<tuple<int, int, int>> h, v;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    if (x1 == x2)
      v.emplace_back(x1, min(y1, y2), max(y1, y2));
    else
      h.emplace_back(y1, min(x1, x2), max(x1, x2));
  }
  int N = h.size(), M = v.size();
  vector<BitSet> bs(N, BitSet(M));
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < M; ++j) {
      int x1, x2, x3;
      int y1, y2, y3;
      tie(y1, x2, x3) = h[i];
      tie(x1, y2, y3) = v[j];
      if (x2 <= x1 && x1 <= x3 && y2 <= y1 && y1 <= y3) bs[i].set(j);
    }
  ll ans = 0;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < i; ++j) {
      int same = (bs[i] & bs[j]).count();
      ans += same * (same - 1) / 2;
    }
  cout << ans << endl;
  return 0;
}

```