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;
struct point{
	double x,y;
	point(){}
	point(double _x,double _y){
		x=_x;y=_y;
	}
	point operator +(const point &b) const{
		return point(x+b.x,y+b.y);
	}
	point operator -(const point &b) const{
		return point(x-b.x,y-b.y);
	}
	double operator ^(const point &b) const{
		return (x*b.x)+(y*b.y);
	}
	double operator *(const point &b) const{
		return x*b.y-y*b.x;
	}
	point operator *(const double &b) const{
		return point(x*b,y*b);
	}
};
point a[105];
int main(){
	int n;
	cin>>n;
	for(int i=0;i<n;++i)
		cin>>a[i].x>>a[i].y;
	double ans=0;
	for(int i=0;i<n;++i)
		if((a[(i+1)%n]-a[i])*(a[(i+2)%n]-a[i])<0)puts("0"),exit(0);
	puts("1");
	return 0;
} 
