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;
const double EPS = 1e-8;
bool dcmp(double d) {
  if (fabs(d) < EPS) {
    return 0;
  } else {
    return d < 0 ? -1 : 1;
  }
}
struct Vector {
  double x, y;
  Vector(double x = 0, double y = 0) : x(x), y(y) {}
};
Vector operator-(const Vector& a, const Vector& b) {
  return Vector(a.x - b.x, a.y - b.y);
}
double cross(const Vector& a, const Vector& b) { return a.x * b.y - a.y * b.x; }
double ccw(const Vector& a, const Vector& b, const Vector& c) {
  return cross(b - a, c - a);
}
int main() {
  int n;
  cin >> n;
  vector<Vector> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
  }
  bool isConvex = true;
  for (int i = 1; i < n - 1; i++) {
    double area = ccw(p[0], p[i], p[i + 1]);
    if (dcmp(area) <= 0) {
      isConvex = false;
    }
  }
  cout << isConvex << endl;
  return 0;
}
