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;
static const double EPS = 1e-10;
static const int COUNTER_CLOCKWISE = 1;
static const int CLOCKWISE = -1;
static const int ONLINE_BACK = 2;
static const int ONLINE_FRONT = -2;
static const int ON_SEGMENT = 0;
class Point {
 public:
  int x, y;
  Point(int x = 0, int y = 0) : x(x), y(y) {}
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
};
double norm(Point a) { return a.x * a.x + a.y * a.y; }
double dot(Point a, Point b) { return a.x * b.x + a.y * b.y; }
double cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
int ccw(Point p0, Point p1, Point p2) {
  Point a = p1 - p0;
  Point b = p2 - p0;
  if (cross(a, b) > EPS) return COUNTER_CLOCKWISE;
  if (cross(a, b) < -EPS) return CLOCKWISE;
  if (dot(a, b) < -EPS) return ONLINE_BACK;
  if (norm(a) < norm(b)) return ONLINE_FRONT;
  return ON_SEGMENT;
}
bool solve(vector<Point> s) {
  if (s.size() <= 3) return true;
  int dir = ccw(s[0], s[1], s[2]);
  for (unsigned int i = 3; i < s.size(); ++i) {
    int res = ccw(s[i - 2], s[i - 1], s[i]);
    if (res != dir && res != ONLINE_BACK) return false;
  }
  return true;
}
int main(int argc, char** argv) {
  vector<Point> s;
  Point p;
  int n;
  scanf("%d", &n);
  while (n--) {
    scanf("%d %d", &p.x, &p.y);
    s.push_back(p);
  }
  printf("%d\n", solve(s) ? 1 : 0);
  return 0;
}
