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 <bits/stdc++.h>
using namespace std;

const double EPS = 1e-8;
const double INF = 1e12;

//point
typedef complex<double> P;
namespace std {
	bool operator < (const P& a, const P& b) {
		return real(a) != real(b) ? real(a) < real(b) : imag(a) < imag(b);
	}
}
double cross(const P& a, const P& b) {
	return imag(conj(a)*b);
}
double dot(const P& a, const P& b) {
	return real(conj(a)*b);
}
int ccw(P a, P b, P 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
}

// line
struct L : public vector<P> {
	L(const P& a, const P& b) {
		push_back(a); push_back(b);
	}
};

// polygon
typedef vector<P> G;

P extreme(const vector<P> &po, const L &l) {
  int k = 0;
  for (int i = 1; i < po.size(); ++i)
    if (dot(po[i], l[1]-l[0]) > dot(po[k], l[1]-l[0])) k = i;
  return po[k];
}

enum { OUT, ON, IN };
int contains(const G& po, const P& p) {
  bool in = false;
  for (int i = 0; i < po.size(); ++i) {
  	P a = po[i] - p, b = po[(i+1)%po.size()] - p;
    if (imag(a) > imag(b)) swap(a, b);
    if (imag(a) <= 0 && 0 < imag(b))
      if (cross(a, b) < 0) in = !in;
    if (cross(a, b) == 0 && dot(a, b) <= 0) return ON;
  }
  return in ? IN : OUT;
}

double area2(const G& po) {
	double A = 0;
  	for (int i = 0; i < po.size(); ++i) 
    	A += cross(po[i], po[(i+1)%po.size()]);
	//????????????????????????????????¨????????????
  	return A/2;
}

bool isconvex(const G &p) {
	int n = p.size();
	if(cross(p[0]-p[n-1],p[n-2]-p[n-1]) < 0) return false;
	for(int i = 1; i < n-1; ++i) {
		if(cross(p[i+1]-p[i],p[i-1]-p[i]) < 0) return false;
	}
	return true;

}
	
int main() {
	int n;
	cin >> n;
	G po;
	for(int i=0; i<n; ++i) {
		double x, y;
		cin >> x >> y;
		P p(x, y);
		po.push_back(p);
	}
	if(isconvex(po)) cout << 1 << endl;
	else cout << 0 << endl;
	return 0;
}