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 <complex>

using namespace std;

const double eps = 1e-10;
typedef complex<double> Point;

inline double cross(const Point& a, const Point& b){
    return imag(conj(a) * b);
}

inline double dot(const Point& a, const Point& b){
    return real(conj(a) * b);
}

int ccw(Point a, Point b, Point c){
    b -= a; c -= a;
    if(cross(b, c) > eps) return  1;
    if(cross(b, c) < -eps)return -1;
    if(dot(b, c) < 0)     return  2;
    if(norm(b) < norm(c)) return -2;
    return 0;
}

typedef vector<Point> Polygon;

bool isconvex(const Polygon& g){
    int n = (int)g.size();
    for(int i=0; i<n; ++i){
        if(ccw(g[i], g[(i+1) % n], g[(i+2) % n]) == -1)return false;
    }
    return true;
}

int main(){
    int n;
    cin >> n;
    Polygon g;
    while(n--){
        int x, y;
        cin >> x >> y;
        g.emplace_back(x, y);
    }
    cout << (isconvex(g)? 1: 0) << '\n';
    return 0;
}