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;
double dot(complex<double> v1, complex<double> v2) {
  return (v1 * conj(v2)).real();
}
double cross(complex<double> v1, complex<double> v2) {
  return (v1 * conj(v2)).imag();
}
enum { COUNTER_CLOCKWISE, CLOCKWISE, ONLINE_FRONT, ONLINE_BACK, ON_SEGMENT };
int ccw(complex<double> p0, complex<double> p1, complex<double> p2) {
  complex<double> a = p1 - p0;
  complex<double> b = p2 - p0;
  if (cross(a, b) > (1e-10)) return CLOCKWISE;
  if (cross(a, b) < -(1e-10)) return COUNTER_CLOCKWISE;
  if (dot(a, b) < -(1e-10)) return ONLINE_BACK;
  if (norm(a) < norm(b)) return ONLINE_FRONT;
  return ON_SEGMENT;
}
int main() {
  double x, y;
  vector<complex<double> > g;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    g.push_back(complex<double>(x, y));
  }
  for (int i = 0; i < n; i++) {
    if (ccw(((g)[((i)-1 + (g).size()) % (g).size()]), ((g)[(i) % (g).size()]),
            ((g)[((i) + 1 + (g).size()) % (g).size()])) != COUNTER_CLOCKWISE) {
      cout << 0 << endl;
      return 0;
    }
  }
  cout << 1 << endl;
  return 0;
}
