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 int INF = INT_MAX / 3;
double EPS = 1e-10;
double add(double a, double b) {
  if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
  return a + b;
}
struct P {
  double x, y;
  P() {}
  P(double x, double y) : x(x), y(y) {}
  P operator+(P p) { return P(add(x, p.x), add(y, p.y)); }
  P operator-(P p) { return P(add(x, -p.x), add(y, -p.y)); }
  P operator*(double d) { return P(d * x, d * y); }
  double dot(P p) { return add(x * p.x, y * p.y); }
  double det(P p) { return add(x * p.y, -y * p.x); }
  void print() {
    printf("%.10lf %.10lf \n", x, y);
    return;
  }
};
int IsConvex(vector<P> &polygon) {
  int n = polygon.size();
  int d = 1;
  P p1, p2;
  for (int i = 0; i < n - 2; i++) {
    p1 = polygon[i + 1] - polygon[i];
    p2 = polygon[i + 2] - polygon[i + 1];
    if (p1.det(p2) <= 0) {
      d = 0;
      break;
    }
  }
  p1 = polygon[n - 1] - polygon[n - 2];
  p2 = polygon[0] - polygon[n - 1];
  if (p1.det(p2) <= 0) {
    d = 0;
  }
  p1 = polygon[0] - polygon[n - 1];
  p2 = polygon[1] - polygon[0];
  if (p1.det(p2) <= 0) {
    d = 0;
  }
  return d;
}
int main() {
  int n;
  cin >> n;
  vector<P> polygon;
  for (int i = 0; i < (int)n; i++) {
    P p;
    cin >> p.x >> p.y;
    polygon.push_back(p);
  }
  cout << IsConvex(polygon) << endl;
  return 0;
}
