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<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
ll read()
{
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int n;
struct Point{double x,y;}p[105];
Point operator -(Point a,Point b){return (Point){a.x-b.x,a.y-b.y};}
double cross(Point a,Point b){return a.x*b.y-a.y*b.x;}
void read(Point &a){a.x=read(),a.y=read();}
bool convex()
{
	for(int i=0;i<n;i++)
	{
		int x=(i+1)%n,y=(i+2)%n;
		if(cross(p[x]-p[i],p[y]-p[x])<0)return 0;
	}
	return 1;
}
int main()
{
	n=read();
	for(int i=0;i<n;i++)read(p[i]);
	printf("%d\n",convex());
	return 0;
}

