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 <fstream>
#include <string>
#include <iomanip>
#include <math.h>
#include <stdbool.h>
#include <algorithm>
using namespace std;

class Point {
    public:
    double x, y;

    Point(double x = 0, double y = 0): 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 a) { return Point(a*x, a*y); }
    Point operator / (double a) { return Point(x/a, y/a); }

    double abs() {return sqrt(norm()); }
    double norm() {return x*x + y*y; }
};

typedef Point Vector;


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;
}

int main(){
    #if 0
    std::ifstream in("input.txt");
    std::cin.rdbuf(in.rdbuf());
    #endif

    int n;
    cin >> n;

    Point i1, i2, p1, p2, p3;
    int isConvex = 1;

    for(int i=0; i<n; i++){
        if(i==0)
        {
            cin >> i1.x >> i1.y;
            cin >> i2.x >> i2.y;
            p1 = i1;
            p2 = i2;
        }
        else
        {
            p1 = p2;
            p2 = p3;
        }
        if(i==n-2){
            p3 = i1;
        }
        else if(i==n-1){
            p3 = i2;
        }
        else{
            cin >> p3.x >> p3.y;
        }

        //cout << p1.x << " " << p1.y << " " << p2.x << " " << p2.y << " " << p3.x << " " << p3.y << "\n";
        
        if( cross(p2-p1,p3-p2) < 0 ){
            isConvex = 0;
            break;
        }
    }

    cout << isConvex <<"\n";
    return 0;
}

