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 eps 1e-9
#define nmax 200
#define f(c,a,b) for(int c=a; c<=b; c++)

using namespace std;
typedef double db;
struct P{
    db x, y;
    P(){}
    P(db x, db y) : x(x) , y(y) {}
    P operator - (P a){ return P(x-a.x, y-a.y); }
    db times(P a){ return x*a.y-y*a.x; }
}po[nmax];
typedef P V;
int n;

bool isconvex(){
    po[0] = po[n];
    f(i,2,n) if( (po[i]-po[i-1]).times(po[i-1]-po[i-2]) > eps ) return false;
    if( (po[1]-po[n]).times(po[n]-po[n-1]) > eps ) return false;
    return true;
}

int main(){
    //freopen("owo.in","r",stdin);
    cin >> n;
    f(i,1,n) scanf("%lf%lf", &po[i].x, &po[i].y);
    printf("%d\n", isconvex());
    return 0;
}
