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>
using namespace std;
typedef long long ll;
//template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
//template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
 
/* attention
    long longのシフト演算には気をつけよう
    タイポした時のデバッグが死ぬほどきつくなるので変数名は最低3字くらい使った方がいいかも
*/
 
#define EPS 0.0000000001
#define equals(a, b) (fabs((a)-(b)) < EPS)
 
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 fabs(x - p.x) < EPS && fabs(y - p.y) < EPS;
    }
};
 
typedef Point Vector;

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

// 多角形
typedef vector<Point> Polygon;

struct Segment{
    Point p1, p2;
};

typedef Segment Line;
 
// 内積
double dot(Vector a, Vector b){
    return a.x*b.x + a.y*b.y;
}
 
// 外積
double cross(Vector a, Vector b){
    return a.x*b.y - a.y*b.x;
}
 
// ベクトルの直交判定
bool isOrthogonal(Vector a, Vector b){
    return equals(dot(a,b), 0.0);
}
 
// ベクトルの平行判定
bool isParallel(Vector a, Vector b){
    return equals(cross(a,b), 0.0);
}
 
// 射影
Point project(Segment s, Point p){
    Vector base = s.p2-s.p1;
    Vector hypo = p-s.p1;
    double t = dot(hypo, base)/base.norm();
    return s.p1 + base*t;
}

// 反射
Point reflect(Segment s, Point p){
    return p + (project(s, p)-p)*2;
}

// 反時計回りCCW
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;

int ccw(Point p0, Point p1, Point p2){
    Vector a = p1-p0;
    Vector b = p2-p0;
    if(cross(a,b) > EPS) return COUNTER_CLOCKWISE;
    if(cross(a,b) < -EPS) return CLOCKWISE;
    if(dot(a, b) < -EPS) return ONLINE_BACK;
    if(a.norm() < b.norm()) return ONLINE_FRONT;

    return ON_SEGMENT;
}

bool intersect(Point p1, Point p2, Point p3, Point p4){
    return (ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
            ccw(p3, p4, p1) * ccw(p3, p4 ,p2) <= 0);
}

bool intersect(Segment s1, Segment s2){
    return intersect(s1.p1, s1.p2, s2.p1, s2.p2);
}

// 2点間の距離
double getDistance(Point a, Point b){
    Vector v = b-a;
    return v.abs();
}

// 点と直線の距離
double getDistanceLP(Line l, Point p){
    return abs(cross(l.p2 - l.p1, p - l.p1) / (l.p2 - l.p1).abs());
}

// 点と線分の距離
double getDistanceSP(Segment s, Point p){
    if(dot(s.p2-s.p1, p-s.p1) < 0.0) return (p-s.p1).abs();
    if(dot(s.p1-s.p2, p-s.p2) < 0.0) return (p-s.p2).abs();
    return getDistanceLP(s, p); 
}

// 線分と線分の距離
double getDistance(Segment s1, Segment s2){
    if(intersect(s1, s2)) return 0.0;
    return min(min(getDistanceSP(s1, s2.p1), getDistanceSP(s1, s2.p2)),
               min(getDistanceSP(s2, s1.p1), getDistanceSP(s2, s1.p2)));
}

// 線分と線分の交点
Point getCrossPoint(Segment s1, Segment s2){
    Vector base = s2.p2 - s2.p1;
    double d1 = abs(cross(base, s1.p1 - s2.p1));
    double d2 = abs(cross(base, s1.p2 - s2.p1));
    double t = d1 / (d1 + d2);
    return s1.p1 + (s1.p2 - s1.p1) * t;
}

// 円と直線の交点
pair<Point, Point> getCrossPoints(Circle c, Line l){
    //必要に応じて交点を持つかチェックして
    Vector pr = project(l, c.c);
    Vector e = (l.p2 - l.p1) / (l.p2 - l.p1).abs();
    double base = sqrt(c.r*c.r - (pr - c.c).norm());
    return make_pair(pr + e*base, pr - e*base);
}

// 円と円の交点
double arg(Vector p) { return atan2(p.y, p.x); }
Vector polar(double a, double r) { return Point(cos(r)*a, sin(r)*a); }

pair<Point, Point> getCrossPoints(Circle c1, Circle c2){
    //必要に応じて交点を持つかチェックして
    double d = (c1.c - c2.c).abs();
    double a = acos((c1.r*c1.r + d*d - c2.r*c2.r) / (2*c1.r*d));
    double t = arg(c2.c - c1.c);
    return make_pair(c1.c + polar(c1.r, t+a), c1.c + polar(c1.r, t-a));
}

// 多角形gに点pが含まれるか
int contains(Polygon g, Point p){
    int n = g.size();
    bool x = false;
    for(int i=0; i<n; i++){
        Point a = g[i]-p, b = g[(i+1)%n]-p;
        if(abs(cross(a,b)) < EPS && dot(a,b) < EPS) return 1; // 辺上
        if(a.y > b.y) swap(a, b);
        if(a.y < EPS && EPS < b.y && cross(a,b) > EPS) x = !x;
    }

    if(x) return 2; // 2はcontain 0はnot contain
    else return 0;
}

// アンドリューのアルゴリズム
Polygon andrewScan( Polygon s ){
    Polygon u, l;
    if( s.size() < 3 ) return s;
    sort(s.begin(), s.end());
    // xが小さいものから2つuに追加
    u.push_back(s[0]);
    u.push_back(s[1]);
    // xが大きいものから2つlに追加
    l.push_back(s[s.size() - 1]);
    l.push_back(s[s.size() - 2]);

    // 凸包の上部を生成
    for( int i=2; i<s.size(); i++ ){
        for( int n=u.size(); n>=2 && ccw(u[n-2], u[n-1], s[i]) == COUNTER_CLOCKWISE; n-- ){ // 辺上の点を含めない場合は　!= CLOCKWIZE に置き換える
            u.pop_back();
        }
        u.push_back(s[i]);
    }

    // 凸包の下部を生成
    for( int i=s.size()-3; i>=0; i--){
        for( int n=l.size(); n>=2 && ccw(l[n-2], l[n-1], s[i]) == COUNTER_CLOCKWISE; n-- ){ // 辺上の点を含めない場合は　!= CLOCKWIZE に置き換える
            l.pop_back();
        }
        l.push_back(s[i]);
    }

    // 時計回りになるように凸包の点の列を生成
    reverse(l.begin(), l.end());
    for( int i=u.size()-2; i>=1; i-- ) l.push_back(u[i]);

    return l;
}

// 平面捜査で線分の交差を検出するプログラム
// 端点の種類
#define BOTTOM 0
#define LEFT 1
#define RIGHT 2
#define TOP 3

class EndPoint {
    public:
    Point p;
    int seg, st; // 入力線分のID、端点の種類
    EndPoint() {}
    EndPoint(Point p, int seg, int st): p(p), seg(seg), st(st) {}

    bool operator < (const EndPoint &ep) const {
        // y座標が小さい順に整列
        if(p.y == ep.p.y){
            return st < ep.st; // yが同一の場合は、下端点、左端点、右端点、上端点の順に並べる
        } else return p.y < ep.p.y;
    }
};

EndPoint EP[2*100000]; // 端点のリスト

// 線分交差問題：マンハッタン幾何
int manhattanIntersection(vector<Segment> S){
    int n = S.size();

    for(int i=0, k=0; i<n; i++){
        // 端点p1, p2が左下を基準に並ぶように調整
        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;           // 二分探索木
    BT.insert(1000000001); // 番兵
    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){
            auto b = BT.lower_bound(S[EP[i].seg].p1.x);
            auto 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, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
        Point p1(x1, y1), p2(x2, y2);
        Segment s;
        s.p1 = p1, s.p2 = p2;
        S.push_back(s);
    }

    cout << manhattanIntersection(S) << endl;
}
