#include <bits/stdc++.h>
using namespace std;
const int N = 10100;
const int delta = 5001;
struct Point {
  int x, y, id;
  Point() {}
  Point(int x, int y, int id) : x(x), y(y), id(id) {}
  inline bool operator<(const Point &b) const { return id < b.id; }
};
int t[N];
const int maxn = 10010;
void Add(int x, int val) {
  while (x <= maxn) {
    t[x] += val;
    x += x & -x;
  }
}
int Ask(int x) {
  int res = 0;
  while (x) {
    res += t[x];
    x -= x & -x;
  }
  return res;
}
int Query(int l, int r) { return Ask(r) - Ask(l - 1); }
long long Calc(long long x) { return x * (x - 1) / 2; }
struct cmp {
  bool operator()(const Point &a, const Point &b) { return a.y > b.y; }
};
vector<Point> H, V;
priority_queue<Point, vector<Point>, cmp> tmp;
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v, x, y; i <= n; ++i) {
    scanf("%d%d%d%d", &u, &v, &x, &y);
    u += delta;
    v += delta;
    x += delta;
    y += delta;
    if (u == x) V.push_back(Point(min(v, y), max(v, y), x));
    if (v == y) H.push_back(Point(min(u, x), max(u, x), y));
  }
  sort(H.begin(), H.end());
  int szH = H.size(), szV = V.size();
  long long ans = 0;
  for (int i = 0; i < szH; ++i) {
    for (int j = 0; j < szV; ++j)
      if (H[i].x <= V[j].id && V[j].id <= H[i].y && V[j].x <= H[i].id &&
          H[i].id <= V[j].y)
        Add(V[j].id, 1), tmp.push(V[j]);
    for (int j = i + 1; j < szH; ++j) {
      while (!tmp.empty() && H[j].id > tmp.top().y) {
        Add(tmp.top().id, -1);
        tmp.pop();
      }
      ans += Calc(Query(H[j].x, H[j].y));
    }
    while (!tmp.empty()) {
      Add(tmp.top().id, -1);
      tmp.pop();
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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 = 10100;
const int delta = 5001;
struct Point {
  int x, y, id;
  Point() {}
  Point(int x, int y, int id) : x(x), y(y), id(id) {}
  inline bool operator<(const Point &b) const { return id < b.id; }
};
int t[N];
const int maxn = 10010;
void Add(int x, int val) {
  while (x <= maxn) {
    t[x] += val;
    x += x & -x;
  }
}
int Ask(int x) {
  int res = 0;
  while (x) {
    res += t[x];
    x -= x & -x;
  }
  return res;
}
int Query(int l, int r) { return Ask(r) - Ask(l - 1); }
long long Calc(long long x) { return x * (x - 1) / 2; }
struct cmp {
  bool operator()(const Point &a, const Point &b) { return a.y > b.y; }
};
vector<Point> H, V;
priority_queue<Point, vector<Point>, cmp> tmp;
int n;
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v, x, y; i <= n; ++i) {
    scanf("%d%d%d%d", &u, &v, &x, &y);
    u += delta;
    v += delta;
    x += delta;
    y += delta;
    if (u == x) V.push_back(Point(min(v, y), max(v, y), x));
    if (v == y) H.push_back(Point(min(u, x), max(u, x), y));
  }
  sort(H.begin(), H.end());
  int szH = H.size(), szV = V.size();
  long long ans = 0;
  for (int i = 0; i < szH; ++i) {
    for (int j = 0; j < szV; ++j)
      if (H[i].x <= V[j].id && V[j].id <= H[i].y && V[j].x <= H[i].id &&
          H[i].id <= V[j].y)
        Add(V[j].id, 1), tmp.push(V[j]);
    for (int j = i + 1; j < szH; ++j) {
      while (!tmp.empty() && H[j].id > tmp.top().y) {
        Add(tmp.top().id, -1);
        tmp.pop();
      }
      ans += Calc(Query(H[j].x, H[j].y));
    }
    while (!tmp.empty()) {
      Add(tmp.top().id, -1);
      tmp.pop();
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```