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 <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAX_N = 100000;
const int INF = 1111111111;

class Point {
public:
    int x, y;

    Point(int x = 0, int 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());
    }
    int 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 x == p.x && y == p.y;
    }
};

typedef Point Vector;

class Segment {
public:
    Point p1, p2;

    Segment(Point p1, Point p2): p1(p1), p2(p2) {}
};

static const int BOTTOM = 0;
static const int LEFT = 1;
static const int RIGHT = 2;
static const int 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 * MAX_N];

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(INF);
    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 = lower_bound(BT.begin(), BT.end(), S[EP[i].seg].p1.x);
            auto e = upper_bound(BT.begin(), BT.end(), S[EP[i].seg].p2.x);
            cnt += distance(b, e);
        }
    }

    return cnt;
}

int main(void) {
    int n; cin >> n;

    int x1, y1, x2, y2;
    vector<Segment> S;
    for (int i = 0; i < n; i++) {
        cin >> x1 >> y1 >> x2 >> y2;
        Segment s(Point(x1, y1), Point(x2, y2));
        S.push_back(s);
    }

    cout << manhattanIntersection(S) << endl;

    return 0;
}