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;
typedef complex<double> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> Poly; 
const double EPS = 1e-10;

// 内積 |a||b|cosθ
double dot(Point a, Point b){
	// return a.real() * b.real() + a.imag() * b.imag();
	return (conj(a) * b).real();
}
// 外積、符号付面積の２倍　|a||b|sinθ
double cross(Point a, Point b){
	// return a.real() * b.imag() - a.imag() * b.real();
	return (conj(a) * b).imag();
}
// 点の位置関係　2点a,bから見た点cの位置関係
int ccw(Point a, Point b, Point c){
    // COUNTER_CLOCKWISE
    if(cross(b - a, c - a) > EPS) return 1;
    // CLOCKWISE
    if(cross(b - a, c - a) < -EPS) return -1;
    // ONLINE_BACK (c--a--b on line)
    if(dot(b - a, c - a) < -EPS) return 2;
    // ONLINE_FRONT (a--b--c on line)
    if(abs(b - a) + EPS < abs(c - a)) return -2;
    // ON_SEGMENT
    return 0;
}

bool isConvex(Poly p){
    int n = p.size();
    for(int i = 0; i < n; i++){
        if(ccw(p[(i - 1 + n) % n], p[i], p[(i + 1) % n]) == -1) return false;
    }
    return true;
}

int main(){
    int n;
    cin >> n;
    Poly pl(n);
    for(int i = 0; i < n; i++){
        double x, y;
        cin >> x >> y;
        pl[i] = {x, y};
    }
    cout << isConvex(pl) << endl;
}
