#include <bits/stdc++.h>
using namespace std;
int tree[10010];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    q;
struct punto {
  int x1, y1, x2, y2;
  punto(int x1, int y1, int x2, int y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
};
bool horizontal(punto* p) { return p->y1 == p->y2; }
bool cruza(punto* h, punto* v) {
  return (h->x1 <= v->x1 && v->x1 <= h->x2) &&
         (v->y1 <= h->y1 && h->y1 <= v->y2);
}
bool f(punto* a, punto* b) { return a->x1 < b->x1; }
void add(int k, int x) {
  while (k < 10010) {
    tree[k] += x;
    k += k & -k;
  }
}
long long sum(int k) {
  long long s = 0;
  while (k >= 1) {
    s += tree[k];
    k -= k & -k;
  }
  return s;
}
long long F(punto* a, punto* b) {
  if (a->x1 == b->x1) return 0;
  int C[] = {a->y1, a->y2, b->y1, b->y2};
  sort(C, C + 4);
  long long k = sum(C[2]) - sum(C[1] - 1);
  return (k * (k - 1)) / 2;
}
punto* A[10010];
int main() {
  int n, x, a, b, c, d, y;
  scanf("%d", &n);
  for (x = 0; x < n; x++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    a += 5002;
    b += 5002;
    c += 5002;
    d += 5002;
    punto* p = new punto(a, b, c, d);
    if (horizontal(p)) {
      if (p->x1 > p->x2) swap(p->x1, p->x2);
    } else {
      if (p->y1 > p->y2) swap(p->y1, p->y2);
    }
    A[x] = p;
  }
  sort(A, A + n, f);
  long long s = 0;
  for (x = 0; x < n; x++)
    if (!horizontal(A[x])) {
      y = 0;
      while (!q.empty() || y < n) {
        if (!q.empty() && y < n) {
          if (q.top().first < A[y]->x1) {
            a = q.top().first;
            b = q.top().second;
            q.pop();
            add(b, -1);
          } else {
            if (horizontal(A[y]) && cruza(A[y], A[x])) {
              q.push(make_pair(A[y]->x2, A[y]->y1));
              add(A[y]->y1, 1);
            } else if (!horizontal(A[y]) && y > x)
              s += F(A[x], A[y]);
            y++;
          }
        } else if (!q.empty()) {
          a = q.top().first;
          b = q.top().second;
          q.pop();
          add(b, -1);
        } else if (y < n) {
          bool hor = horizontal(A[y]);
          if (!hor && y > x)
            s += F(A[x], A[y]);
          else if (hor && cruza(A[y], A[x])) {
            q.push(make_pair(A[y]->x2, A[y]->y1));
            add(A[y]->y1, 1);
          }
          y++;
        }
      }
    }
  cout << s << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int tree[10010];
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    q;
struct punto {
  int x1, y1, x2, y2;
  punto(int x1, int y1, int x2, int y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
};
bool horizontal(punto* p) { return p->y1 == p->y2; }
bool cruza(punto* h, punto* v) {
  return (h->x1 <= v->x1 && v->x1 <= h->x2) &&
         (v->y1 <= h->y1 && h->y1 <= v->y2);
}
bool f(punto* a, punto* b) { return a->x1 < b->x1; }
void add(int k, int x) {
  while (k < 10010) {
    tree[k] += x;
    k += k & -k;
  }
}
long long sum(int k) {
  long long s = 0;
  while (k >= 1) {
    s += tree[k];
    k -= k & -k;
  }
  return s;
}
long long F(punto* a, punto* b) {
  if (a->x1 == b->x1) return 0;
  int C[] = {a->y1, a->y2, b->y1, b->y2};
  sort(C, C + 4);
  long long k = sum(C[2]) - sum(C[1] - 1);
  return (k * (k - 1)) / 2;
}
punto* A[10010];
int main() {
  int n, x, a, b, c, d, y;
  scanf("%d", &n);
  for (x = 0; x < n; x++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    a += 5002;
    b += 5002;
    c += 5002;
    d += 5002;
    punto* p = new punto(a, b, c, d);
    if (horizontal(p)) {
      if (p->x1 > p->x2) swap(p->x1, p->x2);
    } else {
      if (p->y1 > p->y2) swap(p->y1, p->y2);
    }
    A[x] = p;
  }
  sort(A, A + n, f);
  long long s = 0;
  for (x = 0; x < n; x++)
    if (!horizontal(A[x])) {
      y = 0;
      while (!q.empty() || y < n) {
        if (!q.empty() && y < n) {
          if (q.top().first < A[y]->x1) {
            a = q.top().first;
            b = q.top().second;
            q.pop();
            add(b, -1);
          } else {
            if (horizontal(A[y]) && cruza(A[y], A[x])) {
              q.push(make_pair(A[y]->x2, A[y]->y1));
              add(A[y]->y1, 1);
            } else if (!horizontal(A[y]) && y > x)
              s += F(A[x], A[y]);
            y++;
          }
        } else if (!q.empty()) {
          a = q.top().first;
          b = q.top().second;
          q.pop();
          add(b, -1);
        } else if (y < n) {
          bool hor = horizontal(A[y]);
          if (!hor && y > x)
            s += F(A[x], A[y]);
          else if (hor && cruza(A[y], A[x])) {
            q.push(make_pair(A[y]->x2, A[y]->y1));
            add(A[y]->y1, 1);
          }
          y++;
        }
      }
    }
  cout << s << endl;
  return 0;
}

```