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;

#define EPS 1e-10

struct Point{
    double x,y;

    Point(){}
    Point(double x,double y) : x(x),y(y) {}

    Point operator - (const Point &p)const{ return Point(x-p.x,y-p.y); }
};

double dot(const Point &a,const Point &b){ return a.x*b.x+a.y*b.y; }
double cross(const Point &a,const Point &b){ return a.x*b.y - b.x*a.y; }
double norm(const Point &p){ return dot(p,p); }

#define COUNTER_CLOCKWISE 1
#define CLOCKWISE -1
#define ONLINE_BACK 2
#define ONLINE_FRONT -2
#define ON_SEGMENT 0
typedef Point Vector;

int ccw(Point p0,Point p1,Point p2){
    Vector a = p1 - p0;
    Vector 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(norm(a) < norm(b)) return ONLINE_FRONT;
    return ON_SEGMENT;
}

typedef vector<Point> Polygon;

#define prev(G,i) (G[(i-1+G.size())%G.size()])
#define curr(G,i) (G[i%G.size()])
#define next(G,i) (G[(i+1)%G.size()])

bool isConvex(const Polygon &p){
    for(int i = 0 ; i < (int)p.size() ; i++){
        if(ccw(prev(p,i),curr(p,i),next(p,i)) == CLOCKWISE){
            return false;
        }
    }
    return true;
}

int main(){
    int N;
    cin >> N;
    Polygon p(N);
    for(int i = 0 ; i < N ; i++){
        cin >> p[i].x >> p[i].y;
    }
    cout << isConvex(p) << endl;
    return 0;
}