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 <stdio.h>
#include <stack>
#include <math.h>

using namespace std;

int main(){
    int n;
    scanf("%d",&n);
    double x[n+1];
    double y[n+1];
    int i;
    for(i=0; i<n; i++){
        scanf("%lf %lf",&x[i],&y[i]);
    }
    x[n]=x[0];
    y[n]=y[0];
    double v_x[n+1];
    double v_y[n+1];
    for(i=0; i<n; i++){
        v_x[i]=x[i+1]-x[i];
        v_y[i]=y[i+1]-y[i];
    }
    v_x[n]=v_x[0];
    v_y[n]=v_y[0];
    int S=1;
    int T;
    for(i=0; i<n; i++){
        if(v_x[i]*v_y[i+1]-v_x[i+1]*v_y[i] >= 0){
            T=1;
        } else {
            T=0;
        }
        S = S*T;
    }
    if(S != 0){
        printf("1\n");
    } else {
        printf("0\n");
    }
    
    return 0;
}
