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 <algorithm>
#include <cstdio>
#include <iostream>
#include <vector>
#include <set>

using namespace std;

struct Point {
	double x, y;
};

struct Segment {
	Point p1, p2; 
};

#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];

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;
	vector<Segment> s;

	scanf("%d", &n);
	for ( int i = 0; i < n; i++ ) {
		Segment x;
		scanf("%lf %lf %lf %lf", &x.p1.x, &x.p1.y, &x.p2.x, &x.p2.y);
		s.push_back(x);
	}

	printf("%d\n", manhattanIntersection(s));
	
	return 0;
}

