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;
struct vec {
  double x;
  double y;
};
struct vec p_to_vec(vec p1, vec p2) {
  vec v;
  v.x = p2.x - p1.x;
  v.y = p2.y - p1.y;
  return v;
}
double inner_p(vec a, vec b) { return a.x * b.x + a.y * b.y; }
double cross_p(vec a, vec b) { return a.x * b.y - a.y * b.x; }
double norm(vec a) { return sqrt(inner_p(a, a)); }
int ccw(vec p1, vec p2, vec p3) {
  vec a, b;
  a = p_to_vec(p1, p2);
  b = p_to_vec(p1, p3);
  if (fabs(cross_p(a, b)) < 0.0001)
    if (inner_p(a, b) < 0)
      return 0;
    else if (norm(a) < norm(b))
      return 0;
    else
      return 0;
  else if (cross_p(a, b) > 0)
    return 2;
  else
    return -2;
}
int is_con(vector<vec> v) {
  int a, b, n = v.size();
  a = ccw(v[n - 2], v[n - 1], v[0]);
  b = ccw(v[n - 1], v[0], v[1]);
  if (a * b <= 0)
    return 0;
  else
    b = a;
  for (int i = 0; i < n - 2; i++) {
    a = ccw(v[i], v[i + 1], v[i + 2]);
    if (a * b <= 0)
      return 0;
    else
      b = a;
  }
  return 1;
}
int main() {
  int n;
  vector<struct vec> v;
  vec temp;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> temp.x >> temp.y;
    v.push_back(temp);
  }
  cout << is_con(v) << endl;
  return 0;
}
