For given $n$ segments which are parallel to X-axis or Y-axis, find the number of intersections of them.

Constraints

* $1 \leq n \leq 100,000$
* $ -1,000,000,000 \leq x_1, y_1, x_2, y_2 \leq 1,000,000,000$
* Two parallel segments never overlap or touch.
* The number of intersections $\leq 1,000,000$

Input

In the first line, the number of segments $n$ is given. In the following $n$ lines, the $i$-th segment is given by coordinates of its end points in the following format:

$x_1 \; y_1 \; x_2 \; y_2$

The coordinates are given in integers.

Output

Print the number of intersections in a line.

Example

Input

6
2 2 2 5
1 3 5 3
4 1 4 4
5 2 7 2
6 1 6 3
6 5 6 7


Output

3
#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
using namespace std;

class Point {
 public:
  double x, y;

  Point(double x = 0, double y = 0): x(x), y(y) {}

  Point operator + (Point p) { return Point(x + p.x, y + p.y); }
  Point operator - (Point p) { return Point(x - p.x, y - p.y); }
  Point operator * (double a) { return Point(a * x, a * y); }
  Point operator / (double a) { return Point(x / a, y / a); }

  double abs() { return sqrt(norm()); }
  double norm() {return x * x + y * y; }

  bool operator < (const Point &p) const {
    return x != p.x ? x < p.x : y < p.y;
  }

  bool operator == (const Point &p) const {
    return fabs(x - p.x) < EPS && fabs(y - p.y) < EPS;
  }

};

typedef Point Vector;

double dot(Vector a, Vector b) {
  return a.x * b.x + a.y * b.y;
}

double cross(Vector a, Vector b) {
  return a.x * b.y - a.y * b.x;
}

struct Segment {
  Point p1, p2;
};

typedef Segment Line;

class Circle {
 public:
  Point c;
  double r;
  Circle(Point c = Point(), double r = 0.0) : c(c), r(r) {}
};

typedef vector<Point> Polygon;


#define BOTTOM 0
#define LEFT 1
#define RIGHT 2
#define TOP 3

class EndPoint {
 public:
  Point p;
  int seg, st;
  EndPoint() {}
  EndPoint(Point p, int seg, int st): p(p), seg(seg), st(st) {}

  bool operator < (const EndPoint &ep) const {
    if (p.y == ep.p.y) {
      return st < ep.st;
    } else return p.y < ep.p.y;
  }
};

EndPoint EP[2 * 100000];

void swap(Point& a, Point& b) {
  Point tmp = a;
  a.x = b.x;
  a.y = b.y;
  b.x = tmp.x;
  b.y = tmp.y;
}

int manhattanIntersection(vector<Segment> S) {
  int n = S.size();

  for (int i = 0, k = 0; i < n; i++) {
    if (S[i].p1.y == S[i].p2.y) {
      if (S[i].p1.x > S[i].p2.x) swap(S[i].p1, S[i].p2);
    } else if (S[i].p1.y > S[i].p2.y) swap(S[i].p1, S[i].p2);

    if (S[i].p1.y == S[i].p2.y) {
      EP[k++] = EndPoint(S[i].p1, i, LEFT);
      EP[k++] = EndPoint(S[i].p2, i, RIGHT);
    } else {
      EP[k++] = EndPoint(S[i].p1, i, BOTTOM);
      EP[k++] = EndPoint(S[i].p2, i, TOP);
    }
  }

  sort(EP, EP + (2 * n));

  set<int> BT;
  BT.insert(1000000001);
  int cnt = 0;

  for (int i = 0; i < 2*n; i++) {
    if (EP[i].st == TOP) {
      BT.erase(EP[i].p.x);
    } else if (EP[i].st == BOTTOM) {
      BT.insert(EP[i].p.x);
    } else if (EP[i].st == LEFT) {
      set<int>::iterator b = BT.lower_bound(S[EP[i].seg].p1.x);
      set<int>::iterator e = BT.upper_bound(S[EP[i].seg].p2.x);
      cnt += distance(b, e);
    }
  }

  return cnt;
}

int main() {
  vector<Segment> S;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    Segment s;
    cin >> s.p1.x >> s.p1.y >> s.p2.x >> s.p2.y;
    S.push_back(s);
  }
  cout << manhattanIntersection(S) << endl;
  return 0;
}

