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;
double dot(complex<double> a, complex<double> b) {
  return (conj(a) * b).real();
}
double cross(complex<double> a, complex<double> b) {
  return (conj(a) * b).imag();
}
int ccw(complex<double> a, complex<double> b, complex<double> c) {
  if (cross(b - a, c - a) > EPS) return 1;
  if (cross(b - a, c - a) < -EPS) return -1;
  if (dot(b - a, c - a) < -EPS) return 2;
  if (abs(b - a) + EPS < abs(c - a)) return -2;
  return 0;
}
bool isConvex(vector<complex<double> > p) {
  int n = p.size();
  for (int i = 0; i < n; i++) {
    if (ccw(p[(i - 1 + n) % n], p[i], p[(i + 1) % n]) < 0) return false;
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  vector<complex<double> > pl(n);
  for (int i = 0; i < n; i++) {
    double x, y;
    cin >> x >> y;
    pl[i] = {x, y};
  }
  cout << isConvex(pl) << endl;
}
