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 <vector>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <climits>

using namespace std;

#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)

//???
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 norm(){return x*x + y*y;};
    double absolute(){return sqrt(norm());};
    
    bool operator < (const Point &p) const{
        return x != p.x ? x < p.x : y < p.y;
    }
    
    bool operator == (const Point &p) const{
        return equals(x, p.x) && equals(y, p.y);
    }
    
};
typedef Point Vector;                               //????????????

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

int main(){
    
    int n;
    Vector a, b;
    bool convex_flag = true;
    
    cin >> n;
    
    Vector *q = new Vector[n];
    
    for (int i = 0; i < n; i++) {
        cin >> q[i].x;
        cin >> q[i].y;
    }
    
    for (int i = 0; i < n; i++) {
        a = q[(i+1)%n] - q[i];
        b = q[(i+2)%n] - q[i];
        
        if (cross(a, b) < 0) {
            convex_flag = false;
            break;
        }
    }
    
    if(convex_flag) printf("1\n");
    else            printf("0\n");

    return 0;
}