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 <iostream>
#include <vector>
#include <cstdio>

using namespace std;

struct Point {
  double x;
  double y;
};

double dot(Point p0, Point p1, Point p2) {
  return (p1.x - p0.x) * (p2.x - p0.x) + (p1.y - p0.y) * (p2.y - p0.y);
}

double cross(Point p0, Point p1, Point p2) {
  return ( p1.x - p0.x ) * ( p2.y - p0.y ) - ( p2.x - p0.x) * (p1.y - p0.y);
}

bool isConvex(vector<Point> p) {
  int n = p.size();
  for(int i = 0; i < n; i++) {
    if( cross(p[ (i + 1) % n], p[i % n], p[ (i + 2) % n]) > 0) {
      return false;
    }
  }
  return true;
}

int main(void) {
  int q;
  cin >> q;
  vector<Point> p;

  for(int i = 0; i < q; i++) {
    Point pi;
    cin >> pi.x >> pi.y;
    p.push_back(pi);
  }

  if(isConvex(p))
    cout << "1" << endl;
  else
    cout << "0" << endl;
  
  
}
