#include <bits/stdc++.h>
using namespace std;
const int N = 10009;
const int D = 5001;
int n;
vector<vector<pair<int, int>>> hs(N, vector<pair<int, int>>());
vector<vector<pair<int, int>>> vs(N, vector<pair<int, int>>());
vector<vector<int>> ds(N, vector<int>());
class FenwickTree {
 public:
  FenwickTree(int n) { arr.assign(n + 1, 0); }
  int rsq(int a) {
    int res = 0;
    for (; a; a -= LSOne(a)) {
      res += arr[a];
    }
    return res;
  }
  int rsq(int a, int b) { return rsq(b) - (a == 1 ? 0 : rsq(a - 1)); }
  void update(int a, int c) {
    for (; a < arr.size(); a += LSOne(a)) {
      arr[a] += c;
    }
  }

 private:
  vector<int> arr;
  int LSOne(int a) { return ((a) & (-a)); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D;
    x2 += D;
    y1 += D;
    y2 += D;
    if (x1 == x2) {
      vs[x1].emplace_back(min(y1, y2), max(y1, y2));
    } else {
      hs[y1].emplace_back(min(x1, x2), max(x1, x2));
    }
  }
  long long res = 0;
  for (int i = 1; i < N; ++i)
    for (auto h : hs[i]) {
      int l = h.first;
      int r = h.second;
      for (int j = 0; j < N; ++j) ds[j].clear();
      FenwickTree ft(N);
      for (int x = l; x <= r; ++x)
        for (auto v : vs[x]) {
          if (v.first <= i && v.second > i) {
            ft.update(x, 1);
            ds[v.second].push_back(x);
          }
        }
      for (int y = i + 1; y < N; y++) {
        for (auto h : hs[y]) {
          int s = ft.rsq(h.first, h.second);
          res += (s * (s - 1)) / 2ll;
        }
        for (int v : ds[y]) {
          ft.update(v, -1);
        }
      }
    }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 10009;
const int D = 5001;
int n;
vector<vector<pair<int, int>>> hs(N, vector<pair<int, int>>());
vector<vector<pair<int, int>>> vs(N, vector<pair<int, int>>());
vector<vector<int>> ds(N, vector<int>());
class FenwickTree {
 public:
  FenwickTree(int n) { arr.assign(n + 1, 0); }
  int rsq(int a) {
    int res = 0;
    for (; a; a -= LSOne(a)) {
      res += arr[a];
    }
    return res;
  }
  int rsq(int a, int b) { return rsq(b) - (a == 1 ? 0 : rsq(a - 1)); }
  void update(int a, int c) {
    for (; a < arr.size(); a += LSOne(a)) {
      arr[a] += c;
    }
  }

 private:
  vector<int> arr;
  int LSOne(int a) { return ((a) & (-a)); }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += D;
    x2 += D;
    y1 += D;
    y2 += D;
    if (x1 == x2) {
      vs[x1].emplace_back(min(y1, y2), max(y1, y2));
    } else {
      hs[y1].emplace_back(min(x1, x2), max(x1, x2));
    }
  }
  long long res = 0;
  for (int i = 1; i < N; ++i)
    for (auto h : hs[i]) {
      int l = h.first;
      int r = h.second;
      for (int j = 0; j < N; ++j) ds[j].clear();
      FenwickTree ft(N);
      for (int x = l; x <= r; ++x)
        for (auto v : vs[x]) {
          if (v.first <= i && v.second > i) {
            ft.update(x, 1);
            ds[v.second].push_back(x);
          }
        }
      for (int y = i + 1; y < N; y++) {
        for (auto h : hs[y]) {
          int s = ft.rsq(h.first, h.second);
          res += (s * (s - 1)) / 2ll;
        }
        for (int v : ds[y]) {
          ft.update(v, -1);
        }
      }
    }
  cout << res << endl;
  return 0;
}

```