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;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
double x[10010],y[10010];
double ans;
int pd;
int n;
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>x[i]>>y[i];
		x[i+n]=x[i],y[i+n]=y[i];
	}
	for(int j=1;j<=n-1;j++){
		double xx=x[j],yy=y[j];
		for(int i=1;i<=2*n;i++){
			x[i]-=xx,y[i]-=yy;
		}
		for(int i=3+j-1;i<=n+j-1;i++){
			if(x[i-1]*y[i]-x[i]*y[i-1]<0) pd=1;
		}
		for(int i=1;i<=2*n;i++){
			x[i]-=xx,y[i]-=yy;
		}
	}
	cout<<(pd^1)<<endl; 
	return 0;
}
