#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 10002;
const int maxn = N;
struct fenwick {
  int A[maxn];
  int query(int i) {
    int sum = 0;
    while (i > 0) sum += A[i], i -= ((i) & -(i));
    return sum;
  }
  void update(int i, int k) {
    if (i < 1) return;
    while (i < maxn) A[i] += k, i += ((i) & -(i));
  }
  void rupd(int l, int r, int v) {
    update(l, v);
    update(r + 1, -v);
  }
} fenw;
void solve() {
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int>> h, v;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 > y2) swap(y1, y2);
    if (x1 > x2) swap(x1, x2);
    if (x1 == x2) {
      h.push_back({{y1, y2}, x1});
    } else {
      v.push_back({{x1, x2}, y1});
    }
  }
  sort((v).begin(), (v).end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         return a.first.second < b.first.second;
       });
  sort((h).begin(), (h).end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         return a.second < b.second;
       });
  long long ans = 0;
  int a = (int)((h).size()), b = (int)((v).size());
  for (int i = 0; i < a; i++) {
    int k = b - 1;
    int x1 = h[i].second;
    for (int j = a - 1; j > i; j--) {
      int x2 = h[j].second;
      int y1 = max(h[i].first.first, h[j].first.first);
      int y2 = min(h[i].first.second, h[j].first.second);
      if (y2 < y1) continue;
      while (k >= 0 && v[k].first.second >= x2) {
        if (v[k].first.first <= x1) {
          fenw.update(v[k].second, 1);
        }
        k--;
      }
      int tmp = fenw.query(y2) - fenw.query(y1 - 1);
      ans += 1ll * tmp * (tmp - 1) / 2;
    }
    memset(fenw.A, 0, sizeof(fenw.A));
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const int N = 10002;
const int maxn = N;
struct fenwick {
  int A[maxn];
  int query(int i) {
    int sum = 0;
    while (i > 0) sum += A[i], i -= ((i) & -(i));
    return sum;
  }
  void update(int i, int k) {
    if (i < 1) return;
    while (i < maxn) A[i] += k, i += ((i) & -(i));
  }
  void rupd(int l, int r, int v) {
    update(l, v);
    update(r + 1, -v);
  }
} fenw;
void solve() {
  int n;
  cin >> n;
  vector<pair<pair<int, int>, int>> h, v;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 > y2) swap(y1, y2);
    if (x1 > x2) swap(x1, x2);
    if (x1 == x2) {
      h.push_back({{y1, y2}, x1});
    } else {
      v.push_back({{x1, x2}, y1});
    }
  }
  sort((v).begin(), (v).end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         return a.first.second < b.first.second;
       });
  sort((h).begin(), (h).end(),
       [&](pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
         return a.second < b.second;
       });
  long long ans = 0;
  int a = (int)((h).size()), b = (int)((v).size());
  for (int i = 0; i < a; i++) {
    int k = b - 1;
    int x1 = h[i].second;
    for (int j = a - 1; j > i; j--) {
      int x2 = h[j].second;
      int y1 = max(h[i].first.first, h[j].first.first);
      int y2 = min(h[i].first.second, h[j].first.second);
      if (y2 < y1) continue;
      while (k >= 0 && v[k].first.second >= x2) {
        if (v[k].first.first <= x1) {
          fenw.update(v[k].second, 1);
        }
        k--;
      }
      int tmp = fenw.query(y2) - fenw.query(y1 - 1);
      ans += 1ll * tmp * (tmp - 1) / 2;
    }
    memset(fenw.A, 0, sizeof(fenw.A));
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}

```