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 <cstdio>
#include <complex>
#include <iostream>
#include <vector>
#include <utility>
#define X real()
#define Y imag()
#define EPS 1e-10
using namespace std;
typedef complex<double> P;
typedef pair<P, P> L;
typedef vector<P> Poly;

double cross(P a, P b){
  return a.X*b.Y-a.Y*b.X;
}

int is_convex(Poly p){
  int n = p.size();
  if(cross(p[0]-p[n-1],p[n-2]-p[n-1]) < 0) return 0;
  for(int i = 1; i < n-1; ++i){
    if(cross(p[i+1]-p[i],p[i-1]-p[i]) < 0) return 0;
  }
  return 1;
}

int main(){
  int n;
  cin >> n;
  double x, y;
  Poly p(n);
  for(int i = 0; i < n; ++i){
    cin >> x >> y;
    p[i] = P(x,y);
  }
  cout << is_convex(p) << endl;
  return 0;
}