#include <bits/stdc++.h>
using namespace std;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 10004;
vector<pair<int, int> > hori[N], vert[N];
struct bit {
  int n;
  vector<int> tree;
  bit(int n) : n(n) { tree.assign(n + 1, 0); }
  int query(int l, int r) { return query(r) - query(l - 1); }
  int query(int r) {
    int s = 0;
    for (; r > 0; r -= (r & (-r))) s += tree[r];
    return s;
  }
  void update(int i, int v) {
    for (; i <= n; i += (i & (-i))) tree[i] += v;
  }
};
const int D = 5001;
pair<int, int> intersection(pair<int, int> a, pair<int, int> b) {
  if (b.first > a.second || a.first > b.second) {
    return make_pair(-1, -1);
  }
  return make_pair(max(a.first, b.first), min(a.second, b.second));
}
vector<int> fin[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += D;
    b += D;
    c += D;
    d += D;
    if (b == d) {
      hori[b].push_back({min(a, c), max(a, c)});
    } else {
      vert[a].push_back({min(b, d), max(b, d)});
    }
  }
  bit t(N);
  auto c2 = [](long long c) -> long long { return ((c * (c - 1)) / 2); };
  long long ans = 0;
  for (int i = 1; i < N; i++) {
    for (auto h : hori[i]) {
      int l = h.first;
      int r = h.second;
      vector<int> pre;
      for (int j = l; j <= r; j++) {
        for (auto& v : vert[j]) {
          if (v.first <= i && v.second > i) {
            fin[v.second + 1].push_back(j);
            pre.push_back(v.second + 1);
            t.update(j, 1);
          }
        }
      }
      for (int j = i + 1; j < N; j++) {
        for (auto& k : fin[j]) {
          t.update(k, -1);
        }
        for (auto& h2 : hori[j]) {
          auto ii = intersection(h, h2);
          if (ii.first != -1) {
            int c = t.query(ii.first, ii.second);
            ans += c2(c);
          }
        }
      }
      for (auto& j : pre) {
        fin[j].clear();
      }
    }
  }
  cout << ans << "\n";
}

### Prompt

Your challenge is to write 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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 10004;
vector<pair<int, int> > hori[N], vert[N];
struct bit {
  int n;
  vector<int> tree;
  bit(int n) : n(n) { tree.assign(n + 1, 0); }
  int query(int l, int r) { return query(r) - query(l - 1); }
  int query(int r) {
    int s = 0;
    for (; r > 0; r -= (r & (-r))) s += tree[r];
    return s;
  }
  void update(int i, int v) {
    for (; i <= n; i += (i & (-i))) tree[i] += v;
  }
};
const int D = 5001;
pair<int, int> intersection(pair<int, int> a, pair<int, int> b) {
  if (b.first > a.second || a.first > b.second) {
    return make_pair(-1, -1);
  }
  return make_pair(max(a.first, b.first), min(a.second, b.second));
}
vector<int> fin[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += D;
    b += D;
    c += D;
    d += D;
    if (b == d) {
      hori[b].push_back({min(a, c), max(a, c)});
    } else {
      vert[a].push_back({min(b, d), max(b, d)});
    }
  }
  bit t(N);
  auto c2 = [](long long c) -> long long { return ((c * (c - 1)) / 2); };
  long long ans = 0;
  for (int i = 1; i < N; i++) {
    for (auto h : hori[i]) {
      int l = h.first;
      int r = h.second;
      vector<int> pre;
      for (int j = l; j <= r; j++) {
        for (auto& v : vert[j]) {
          if (v.first <= i && v.second > i) {
            fin[v.second + 1].push_back(j);
            pre.push_back(v.second + 1);
            t.update(j, 1);
          }
        }
      }
      for (int j = i + 1; j < N; j++) {
        for (auto& k : fin[j]) {
          t.update(k, -1);
        }
        for (auto& h2 : hori[j]) {
          auto ii = intersection(h, h2);
          if (ii.first != -1) {
            int c = t.query(ii.first, ii.second);
            ans += c2(c);
          }
        }
      }
      for (auto& j : pre) {
        fin[j].clear();
      }
    }
  }
  cout << ans << "\n";
}

```