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>
#define Re real()
#define Im imag()
using namespace std;

const double eps = 1e-9;
typedef complex<double> Point;
typedef Point Vector;
Point P[50050];
int n, k;
vector<Point> qs;

bool cmp(Point a, Point b)
{
    if(fabs(a.Re - b.Re) < eps) return a.Im < b.Im;
    else return a.Re < b.Re;
}

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

int convex_hull()
{
    k = 0;
    for(int i = 0; i < n; i++)
    {
        while(k > 1 && cross(qs[k - 1] - qs[k - 2], P[i] - qs[k - 1]) <= 0) k--;
        qs[k++] = P[i];
    }
    for(int i = n - 2, t = k; i >= 0; i--)
    {
        while(k > t && cross(qs[k - 1] - qs[k - 2], P[i] - qs[k - 1]) <= 0) k--;
        qs[k++] = P[i];
    }
    return k - 1;
}

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 0; i < n; i++)
        {
            double tx, ty;
            scanf("%lf%lf", &tx, &ty);
            P[i] = Point(tx, ty);
        }

        int flag = 1;
        for(int i = 0; i < n; i++)
        {
            Vector v1 = P[(i - 1 + n) % n] - P[i];
            Vector v2 = P[(i + 1) % n] - P[i];
            if(cross(v1, v2) > 0) flag = 0;
        }
        printf("%d\n",flag);
    }
    return 0;
}

/*

5
0 0
2 0
2 2
0 2
1 1

*/