For a given polygon g, print "1" if g is a convex polygon, "0" otherwise. Here, in a convex polygon, all interior angles are less than or equal to 180 degrees.

g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n-1) are sides of the polygon. The line segment connecting pn and p1 is also a side of the polygon.

Constraints

* 3 ≤ n ≤ 100
* -10000 ≤ xi, yi ≤ 10000
* No point of the polygon will occur more than once.
* Two sides of the polygon can intersect only at a common endpoint.

Input

g is given by coordinates of the points p1,..., pn in the following format:


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points. The coordinate of a point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them.

Output

Print "1" or "0" in a line.

Examples

Input

4
0 0
3 1
2 3
0 3


Output

1


Input

5
0 0
2 0
1 1
2 2
0 2


Output

0
#include <iostream>
#include <vector>
#include <iomanip>

class Point{
public:
	int x, y;
	Point(int ax = 0, int ay = 0):x(ax),y(ay){};
};

int sub_area(const Point &p1, const Point &p2){
	return (p1.x - p2.x) * (p1.y + p2.y);
}
bool judge(const Point &p1, const Point &p2, const Point &p3){
	return (p1.x * p3.y + p2.x * p1.y + p3.x * p2.y) <= (p1.x * p2.y + p2.x * p3.y + p3.x * p1.y);
}
bool fold(std::vector<Point> vec){
	bool flag = true;
	for (auto i = 0; i < vec.size() and flag; ++i){
		flag = flag && judge(vec.at((i - 1) % vec.size()), vec.at(i % vec.size()), vec.at((i + 1) % vec.size()));
	}
	return flag;
}
int main(){
	int n;
	std::cin >> n;
	std::vector<Point> vec;
	for (auto i = 0; i < n; ++i){
		int x, y;
		std::cin >> x >> y;
		vec.push_back(Point(x, y));
	}
	if (fold(vec)){
		std::cout << 1 << std::endl;
	}else{
		std::cout << 0 << std::endl;
	}
}