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 double ll;

struct Point
{
    ll x,y;
    Point(){ x = y = 0.0; }
    Point(ll _x, ll _y) : x(_x), y(_y){}

    Point operator+ (Point p)
    {
        return Point(p.x + x, p.y + y);
    }

    Point operator- (Point p)
    {
        return Point(x - p.x, y - p.y);
    }

    ll operator* (Point p)
    {
        return p.x * x + p.y * y;
    }

    Point operator* (ll t)
    {
        return Point(x*t, y*t);
    }

    ll operator~ ()
    {
        return x * x + y * y;
    }

    ll operator% (Point p)
    {
        return x * p.y - y * p.x;
    }

};

Point polygon[200];
int n;

bool ccw(Point a, Point b, Point p)
{
    return ( (b-a)%(p-a) ) < 0;
}

bool isConvex()
{
    if(n+1 <= 3) return false;
    bool isLeft = ccw(polygon[0], polygon[1], polygon[2]);
    for(int i = 1 ; i < n; i++)
        if(ccw(polygon[i-1],polygon[i],polygon[i+1]) != isLeft)
            return false;
    return true;

}

int main()
{
    cout << fixed;
    cout.precision(1);
    ios::sync_with_stdio(0); cin.tie(0);
    Point p;
    cin >> n;
    for(int i = 0 ; i < n ; i++)
        cin >> polygon[i].x >> polygon[i].y;
    polygon[n] = polygon[0];
    cout << isConvex() << '\n';
    return 0;
}