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>
#define lb long double
#define eps 1e-10
#define N 110
using namespace std;
struct Po{lb x,y;}a[N];
int n;
void get(Po &x){scanf("%Lf %Lf",&x.x,&x.y);}
Po operator -(Po x,Po y){return (Po){x.x-y.x,x.y-y.y};}
lb operator *(Po x,Po y){return x.x*y.y-x.y*y.x;}
int pa(int x){if (x<0)x+=n;if (x>=n)x-=n;return x;}
int main(){
	scanf("%d",&n);
	for (int i=0;i<n;i++)get(a[i]);
	for (int i=0;i<n;i++)if ((a[pa(i+1)]-a[i])*(a[pa(i-1)]-a[i])<-eps){puts("0");return 0;}
	puts("1");
	return 0;
}
