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 <iostream>
#include <complex>
#include <utility>
#include <vector>
using namespace std;

typedef complex<double> Point, Vector;
typedef pair<Point, Point> Segment, Line;
typedef vector<Point> Polygon;

#define X real()
#define Y imag()
#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)

double dot(Vector a, Vector b){
    return a.X * b.X + a.Y * b.Y;
}

double cross(Vector a, Vector b){
    return a.X * b.Y - a.Y * b.X;
}

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;

// ベクトルと点の位置関係
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;
}

bool isConvex(Polygon pol){
    int n = pol.size();

    for(int i=0; i<n; i++){
        if(ccw(pol[i], pol[(i+1)%n], pol[(i+2)%n]) == CLOCKWISE){
            return false;
        }
    }

    return true;
}

int main(){
    int n; cin>>n;
    Polygon pol;
    for(int i=0; i<n; i++){
        double x, y; cin>>x>>y;
        pol.emplace_back(x, y);
    }

    cout << isConvex(pol) << endl;
}
