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
#include<bits/stdc++.h>
#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)
#define BOTTOM 0
#define LEFT 1
#define RIGHT 2
#define TOP 3
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(x * a, y * a);}
    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 equals(x, p.x) && equals(y, p.y);
    }
};typedef Point Vector;

struct Segment {
    Point p1, p2;
    Segment(Point p1, Point p2):p1(p1), p2(p2){}
};typedef Segment Line;
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];
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(){
    int n; cin >> n;
    vector<Segment> S;
    for (int i = 0; i < n; i++){
        int x1, x2, y1, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        Segment s = Segment(Point(x1, y1), Point(x2, y2));
        S.push_back(s);
    }
    cout << manhattanIntersection(S) << endl;
}


