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;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define per(i,a,n) for(int i=(n)-1;i>=(a);i--)
#define MP make_pair

typedef double db;

const db EPS = 1e-8;

inline int sign(db a) {
	return a < -EPS ? -1 : a > EPS;
}

inline int cmp(db a, db b){
	return sign(a-b);
}

struct P {
	db x, y;
	P() {
	}
	P(db _x, db _y) :
			x(_x), y(_y) {
	}
	P operator+(P p) {
		return P(x + p.x, y + p.y);
	}
	P operator-(P p)  {
		return P(x - p.x, y - p.y);
	}
	P operator*(db d) {
		return P(x * d, y * d);
	}
	P operator/(db d) {
		return P(x / d, y / d);
	}
	bool operator<(P p) const {
		int c = sign(x - p.x);
		if (c)
			return c == -1;
		return sign(y - p.y) == -1;
	}
	db dot(P p) {
		return x * p.x + y * p.y;
	}
	db det(P p) {
		return x * p.y - y * p.x;
	}
	db distTo(P p) { return (*this-p).abs(); }

	db alpha() {
		return atan2(y, x);
	}
	void read() {
		cin>>x>>y;
	}
	db abs() {
		return sqrt(abs2());
	}
	db abs2() {
		return x * x + y * y;
	}
	P rot90(){
		return P(-y,x);
	}
};

#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))

P isLL(P p1, P p2, P q1, P q2) {
	db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
	return (p1 * a2 + p2 * a1) / (a1 + a2);
}

bool intersect(db l1,db r1,db l2,db r2){
	if(l1>r1) swap(l1,r1); if(l2>r2) swap(l2,r2); 
	return !( cmp(r1,l2) == -1 || cmp(r2,l1) == -1 );
}

bool isSS(P p1, P p2, P q1, P q2){
    return intersect(p1.x,p2.x,q1.x,q2.x) && intersect(p1.y,p2.y,q1.y,q2.y) && 
    crossOp(p1,p2,q1) * crossOp(p1,p2,q2) <= 0 && crossOp(q1,q2,p1)
            * crossOp(q1,q2,p2) <= 0;
}

bool isMiddle(db a, db m, db b) {
    return sign(a - m) == 0 || sign(b - m) == 0 || (a < m != b < m);
}
 
bool isMiddle(P a, P m, P b) {
    return isMiddle(a.x, m.x, b.x) && isMiddle(a.y, m.y, b.y);
}

bool onSegment(P p1, P p2, P q){
	return crossOp(p1,p2,q) == 0 && isMiddle(p1, q, p2);
}

P proj(P p1, P p2, P q) {
    P dir = p2 - p1;
    return p1 + dir * (dir.dot(q - p1) / dir.abs2());
}

P reflect(P p1, P p2, P q){
	return proj(p1,p2,q) * 2 - q;
}

db nearest(P p1,P p2,P q){
	P h = proj(p1,p2,q);
	if(isMiddle(p1,h,p2))
		return q.distTo(h);
	return min(p1.distTo(q),p2.distTo(q));
}

db disSS(P p1, P p2, P q1, P q2){
	if(isSS(p1,p2,q1,q2)) return 0;
	return min(min(nearest(p1,p2,q1),nearest(p1,p2,q2)), min(nearest(q1,q2,p1),nearest(q1,q2,p2)) );
}

db getrad(P p1,P p2){
	return atan2l(p1.det(p2),p1.dot(p2));
}

db incircle(P p1, P p2, P p3){
	db A = p1.distTo(p2);
	db B = p2.distTo(p3);
	db C = p3.distTo(p1);
	return sqrtl(A*B*C/(A+B+C));
}

db area(vector<P> ps){
	db ret = 0; rep(i,0,ps.size()) ret += ps[i].det(ps[(i+1)%ps.size()]); 
	return abs(ret/2);
}

int main(){
	int n;cin>>n;
	vector<P> ps(n);
	rep(i,0,n) ps[i].read();
	bool ok = 1;
	rep(i,0,n) ok &= crossOp(ps[i],ps[(i+1)%n],ps[(i+2)%n])>=0;
	printf("%d\n",ok);
	return 0;
}