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 <vector>
#include <iomanip>
#include <cmath>

using namespace std;

const double EPS = 1e-8;

int dcmp(double d) {
    if (fabs(d) < EPS) {
        return 0;
    }
    else {
        return d < 0? -1 : 1;
    }
}

struct Vector {
    double x, y;
    Vector(double x = 0, double  y = 0):x(x), y(y) {
    }
};

typedef Vector Point;

Vector operator-(const Vector& a, const Vector& b) {
    return Vector(a.x - b.x, a.y - b.y);
}

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

double ccw(const Point& a, const Point& b, const Point& c) {
    return cross(b - a, c - a);
}

int main() {
    int n; 
    cin >> n;
    vector<Point> p(n);
    for (int i = 0; i < n; i++) {
        cin >> p[i].x >> p[i].y;
    }
    
    bool isConvex = true;
    for (int i = 0; i < n - 1; i++) {
        double area = ccw(p[i], p[(i + 1) % n], p[(i+ 2) % n]);
        if (dcmp(area) < 0) {
            isConvex = false;
            break;
        }
    }
    
    cout << (isConvex? 1 : 0) << endl;
    
    return 0;
}