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 complex<double> qua;
vector<qua> v;
const double epx=1e-7;
int ccw(qua a,qua b,qua c)
{
    b-=a,c-=a,a=c*conj(b);
    if(a.imag()>epx) return 1;
    if(a.imag()<-epx) return -1;
    if(a.real()<-epx) return 2;
    if(abs(b)+epx<abs(c)) return -2;
    return 0;
}
bool judge_convex(vector<qua> &ve)
{
    int len_ve=ve.size();
    int flag1=1,flag2=1;
    for(int i=0;i<len_ve;i++)
    {
        if(ccw(ve[i],ve[(i+1)%len_ve],ve[(i+2)%len_ve])==-1) flag1=0;
        if(ccw(ve[i],ve[(i+1)%len_ve],ve[(i+2)%len_ve])==1) flag2=0;
    }
    return flag1|flag2;
}
int main()
{
    int n;
    double x,y;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf%lf",&x,&y);
        v.push_back(qua(x,y));
    }
    printf("%d\n",judge_convex(v));
}

