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-10;
const double INF = 1e12;
struct L {
  complex<double> a, b, v;
  L() {}
  L(complex<double> _a, complex<double> _b) : a(_a), b(_b), v(b - a) {}
  L(double _ax, double _ay, double _bx, double _by)
      : L(complex<double>(_ax, _ay), complex<double>(_bx, _by)) {}
};
double cross(complex<double> a, complex<double> b) { return imag(conj(a) * b); }
double dot(complex<double> a, complex<double> b) { return real(conj(a) * b); }
long long ccw(complex<double> p0, complex<double> p1, complex<double> p2) {
  if (cross(p1 - p0, p2 - p0) > 0) return +1;
  if (cross(p1 - p0, p2 - p0) < 0) return -1;
  if (dot(p1 - p0, p2 - p0) < 0) return +2;
  if (dot(p0 - p1, p2 - p1) < 0) return -2;
  return 0;
}
bool isConvex(vector<complex<double> > g) {
  long long n = g.size();
  for (long long i = 0; i < (n); i++) {
    if (ccw(g[(i - 1 + g.size()) % g.size()], g[i], g[(i + 1) % g.size()]) <
        0) {
      return false;
    }
  }
  return true;
}
signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  long long n;
  cin >> n;
  vector<complex<double> > g;
  while (n--) {
    double x, y;
    cin >> x >> y;
    g.emplace_back(x, y);
  }
  cout << isConvex(g) << endl;
}
