#include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30;
const long long linf = 1LL << 62;
const int MAX = 1020000;
long long dy[8] = {0, -1, 0, 1, 1, -1, 1, -1};
long long dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const double pi = acos(-1);
const double eps = 1e-7;
template <typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  } else
    return false;
}
template <typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  } else
    return false;
}
template <typename T>
inline void print(T &a) {
  for (long long i = 0; i < (a.size()); i++) cout << a[i] << " ";
  cout << "\n";
}
template <typename T1, typename T2>
inline void print2(T1 a, T2 b) {
  cout << a << " " << b << "\n";
}
template <typename T1, typename T2, typename T3>
inline void print3(T1 a, T2 b, T3 c) {
  cout << a << " " << b << " " << c << "\n";
}
long long pcount(long long x) { return __builtin_popcountll(x); }
const long long mod = 1e9 + 7;
template <typename T>
struct BIT {
  vector<T> dat;
  long long sz;
  BIT(long long sz) : sz(sz) { dat.assign(++sz, 0); }
  T sum(long long k) {
    T ret = 0;
    for (++k; k > 0; k -= k & -k) ret += dat[k];
    return (ret);
  }
  void add(long long k, T x) {
    for (++k; k < dat.size(); k += k & -k) dat[k] += x;
  }
  long long get(T k) {
    if (k <= 0) return 0;
    long long ret = 0;
    int n = 1;
    while (n < sz) n *= 2;
    for (int i = n / 2; i > 0; i /= 2) {
      if (ret + i < sz && dat[ret + i] < k) {
        k -= dat[ret + i];
        ret += i;
      }
    }
    return ret;
  }
};
BIT<long long> bit(10101);
vector<tuple<long long, long long, long long> > x, y;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (n); i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if (a == c) {
      if (b > d) swap(b, d);
      x.push_back({a, b, d});
    } else {
      if (a > c) swap(a, c);
      y.push_back({a, c, b});
    }
  }
  sort((x).begin(), (x).end());
  sort((y).begin(), (y).end());
  long long ans = 0;
  for (long long i = 0; i < (x.size()); i++) {
    int id = 0;
    multiset<pair<int, int> > se;
    int a, b, c;
    tie(a, b, c) = x[i];
    for (long long j = 0; j < (y.size()); j++) {
      int x1, x2, Y;
      tie(x1, x2, Y) = y[j];
      if (x1 > a) break;
      se.emplace(x2, Y);
      bit.add(Y + 5000, 1);
    }
    for (int j = (i + 1); j < (x.size()); j++) {
      int d, e, f;
      tie(d, e, f) = x[j];
      if (a == d) continue;
      while (!se.empty()) {
        pair<long long, long long> p = *se.begin();
        if (p.first < d) {
          bit.add(p.second + 5000, -1);
          se.erase(se.begin());
        } else {
          break;
        }
      }
      if (min(c, f) < max(b, e)) continue;
      long long t = bit.sum(min(c, f) + 5000) - bit.sum(max(b, e) + 4999);
      ans += t * (t - 1) / 2;
    }
    while (!se.empty()) {
      bit.add((*se.begin()).second + 5000, -1);
      se.erase(se.begin());
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t = 1;
  while (t--) {
    solve();
  }
}

### 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;
const int inf = 1 << 30;
const long long linf = 1LL << 62;
const int MAX = 1020000;
long long dy[8] = {0, -1, 0, 1, 1, -1, 1, -1};
long long dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const double pi = acos(-1);
const double eps = 1e-7;
template <typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {
  if (a > b) {
    a = b;
    return true;
  } else
    return false;
}
template <typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {
  if (a < b) {
    a = b;
    return true;
  } else
    return false;
}
template <typename T>
inline void print(T &a) {
  for (long long i = 0; i < (a.size()); i++) cout << a[i] << " ";
  cout << "\n";
}
template <typename T1, typename T2>
inline void print2(T1 a, T2 b) {
  cout << a << " " << b << "\n";
}
template <typename T1, typename T2, typename T3>
inline void print3(T1 a, T2 b, T3 c) {
  cout << a << " " << b << " " << c << "\n";
}
long long pcount(long long x) { return __builtin_popcountll(x); }
const long long mod = 1e9 + 7;
template <typename T>
struct BIT {
  vector<T> dat;
  long long sz;
  BIT(long long sz) : sz(sz) { dat.assign(++sz, 0); }
  T sum(long long k) {
    T ret = 0;
    for (++k; k > 0; k -= k & -k) ret += dat[k];
    return (ret);
  }
  void add(long long k, T x) {
    for (++k; k < dat.size(); k += k & -k) dat[k] += x;
  }
  long long get(T k) {
    if (k <= 0) return 0;
    long long ret = 0;
    int n = 1;
    while (n < sz) n *= 2;
    for (int i = n / 2; i > 0; i /= 2) {
      if (ret + i < sz && dat[ret + i] < k) {
        k -= dat[ret + i];
        ret += i;
      }
    }
    return ret;
  }
};
BIT<long long> bit(10101);
vector<tuple<long long, long long, long long> > x, y;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < (n); i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if (a == c) {
      if (b > d) swap(b, d);
      x.push_back({a, b, d});
    } else {
      if (a > c) swap(a, c);
      y.push_back({a, c, b});
    }
  }
  sort((x).begin(), (x).end());
  sort((y).begin(), (y).end());
  long long ans = 0;
  for (long long i = 0; i < (x.size()); i++) {
    int id = 0;
    multiset<pair<int, int> > se;
    int a, b, c;
    tie(a, b, c) = x[i];
    for (long long j = 0; j < (y.size()); j++) {
      int x1, x2, Y;
      tie(x1, x2, Y) = y[j];
      if (x1 > a) break;
      se.emplace(x2, Y);
      bit.add(Y + 5000, 1);
    }
    for (int j = (i + 1); j < (x.size()); j++) {
      int d, e, f;
      tie(d, e, f) = x[j];
      if (a == d) continue;
      while (!se.empty()) {
        pair<long long, long long> p = *se.begin();
        if (p.first < d) {
          bit.add(p.second + 5000, -1);
          se.erase(se.begin());
        } else {
          break;
        }
      }
      if (min(c, f) < max(b, e)) continue;
      long long t = bit.sum(min(c, f) + 5000) - bit.sum(max(b, e) + 4999);
      ans += t * (t - 1) / 2;
    }
    while (!se.empty()) {
      bit.add((*se.begin()).second + 5000, -1);
      se.erase(se.begin());
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t = 1;
  while (t--) {
    solve();
  }
}

```