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 <complex>
#include <vector>
using namespace std;

typedef complex<double> point;
double cross(const point& a, const point& b) {
	return imag(conj(a)*b);
}
double dot(const point& a, const point& b) {
	return real(conj(a)*b);
}
int ccw(point a, point b, point c) {
	b -= a; c -= a;
	if (cross(b, c) > 0)   return +1;  // counter clockwise
	if (cross(b, c) < 0)   return -1;  // clockwise
	if (dot(b, c) < 0)     return +2;  // c--a--b on line
	if (norm(b) < norm(c)) return -2;  // a--b--c on line
	return 0;  // a--c--b on line or a----bc
}

typedef vector<point> polygon;
bool is_convex(const polygon &P) {
	size_t n = P.size();
	for (int i = 0; i < n; ++i)
		if (ccw(P[(i+n-1) % n], P[i % n], P[(i+1) % n]) == -1) return false;
	return true;
}

int main() {
	int n;
	while (cin >> n) {
		polygon pol(n);
		for (auto& p : pol) {
			int x, y;
			cin >> x >> y;
			p = point(x, y);
		}
		cout << is_convex(pol) << endl;
	}
	return 0;
}