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 EPS 1e-10
#define equals(a,b) (fabs((a)-(b)) < EPS)
#define rep(i,n) for(int i=0;i<n;++i)
typedef long long ll;
using namespace std;

static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;

// 点
struct Point {
    double x,y;
    Point(){}
    Point(double x, double y) : x(x),y(y){}
    Point operator+(Point p) {return Point(x+p.x, y+p.y);}
    Point operator-(Point p) {return Point(x-p.x, y-p.y);}
    Point operator*(double k){return Point(x*k,y*k);}
    Point operator/(double k){return Point(x/k,y/k);}
    double norm(){return x*x+y*y;}
    double abs(){sqrt(norm());}
    bool operator == (const Point &p) const {return equals(x,p.x)&&equals(y,p.y);}
};

typedef Point P;
typedef vector<Point> Polygon;

double cross(Point a, Point b) {return a.x*b.y-a.y*b.x;}

double area(Polygon s){
    double res=0;
    for(int i=0;i<(int)s.size();i++){
        res+=cross(s[i],s[(i+1)%s.size()])/2.0;
    }
    return res;
}

// 内積
double dot(Point a, Point b) {return a.x*b.x + a.y*b.y;}

// 2直線の直行判定
bool is_orthogonal(Point a1, Point a2, Point b1, Point b2) {
    return equals(dot(a1-a2, b1-b2), 0.0);
}
// 2直線の平行判定
bool is_parallel(Point a1, Point a2, Point b1, Point b2) {
    return equals(cross(a1-a2, b1-b2), 0.0);
}
// 点cが直線ab上にあるかないか
bool is_point_on_INF_line(Point a, Point b, Point c) {
    return equals(cross(b-a,c-a), 0.0);
}
// 点cが線分ab上にあるかないか
bool is_point_on_LIMITED_line(Point a, Point b, Point c) {
    return (Point(a-c).abs()+Point(c-b).abs() < Point(a-b).abs() + EPS);
}
// 直線と点の距離
double distance_l_p(Point a, Point b, Point c) {return abs(cross(b-a, c-a)) / (b-a).abs();}

// 点a,bを端点とする線分と点cとの距離
double distance_ls_p(Point a, Point b, Point c) {
    if (dot(b-a, c-a) < EPS) return (c-a).abs();
    if (dot(a-b, c-b) < EPS) return (c-b).abs();
    return abs(cross(b-a, c-a)) / (b-a).abs();
}

// 点が線分のどちら側にあるかを計算
int ccw(Point p0,Point p1,Point p2) {
    P a = p1-p0;
    P b = p2-p0;
    if(cross(a,b) > EPS) return COUNTER_CLOCKWISE;
    if(cross(a,b) < -EPS) return CLOCKWISE;
    if(dot(a,b) < -EPS) return ONLINE_BACK;
    if(a.norm()<b.norm()) return ONLINE_FRONT;
    return ON_SEGMENT;
}

bool isConvex(Polygon p){
    bool f=1;
    int n=p.size();
    for(int i=0;i<n;i++){
        int t=ccw(p[(i+n-1)%n],p[i],p[(i+1)%n]);
        f&=t!=CLOCKWISE;
    }
    return f;
}

int main() {
    int n;cin>>n;
    Polygon pl;
    rep(i,n) {
        double x,y; cin>>x>>y;
        pl.push_back(Point(x,y));
    }
    if (isConvex(pl)) cout << 1 << endl;
    else cout << 0 << endl;
}
