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 <stdio.h>
#include <stack>
#include <math.h>

using namespace std;

struct Point{
	double x,y;
};

int main(){

	int n;
	scanf("%d",&n);
	double gaiseki;

	bool FLG = true;

	Point points[n],left,right;

	for(int i=0; i < n; i++){
		scanf("%lf %lf",&points[i].x,&points[i].y);
	}

	for(int i=0; i < n; i++){
		if(i == 0){
			right.x = points[1].x - points[0].x;
			right.y = points[1].y - points[0].y;

			left.x = points[n-1].x - points[0].x;
			left.y = points[n-1].y - points[0].y;

			gaiseki = right.x*left.y - right.y*left.x;
		}else if(i == n-1){
			right.x = points[0].x - points[n-1].x;
			right.y = points[0].y - points[n-1].y;

			left.x = points[n-2].x - points[n-1].x;
			left.y = points[n-2].y - points[n-1].y;

			gaiseki = right.x*left.y - right.y*left.x;
		}else{
			right.x = points[i+1].x - points[i].x;
			right.y = points[i+1].y - points[i].y;

			left.x = points[i-1].x - points[i].x;
			left.y = points[i-1].y - points[i].y;

			gaiseki = right.x*left.y - right.y*left.x;
		}
		if(gaiseki < 0){
			FLG = false;
			break;
		}
	}

	if(FLG)printf("1\n");
	else{
		printf("0\n");
	}

    return 0;
}