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-9;
complex<double> P[50050];
int n, k;
vector<complex<double> > qs;
bool cmp(complex<double> a, complex<double> b) {
  if (fabs(a.real() - b.real()) < eps)
    return a.imag() < b.imag();
  else
    return a.real() < b.real();
}
double cross(const complex<double>& a, const complex<double>& b) {
  return (conj(a) * b).imag();
}
int convex_hull() {
  k = 0;
  for (int i = 0; i < n; i++) {
    while (k > 1 && cross(qs[k - 1] - qs[k - 2], P[i] - qs[k - 1]) <= 0) k--;
    qs[k++] = P[i];
  }
  for (int i = n - 2, t = k; i >= 0; i--) {
    while (k > t && cross(qs[k - 1] - qs[k - 2], P[i] - qs[k - 1]) <= 0) k--;
    qs[k++] = P[i];
  }
  return k - 1;
}
double area_convex() {
  double ans = 0;
  complex<double> Pivot = qs[0];
  for (int i = 1; i < qs.size(); i++)
    ans += cross(qs[i] - qs[0], qs[i + 1] - qs[0]) / 2.0;
  return ans;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      double tx, ty;
      scanf("%lf%lf", &tx, &ty);
      P[i] = complex<double>(tx, ty);
    }
    sort(P, P + n, cmp);
    qs.resize(2 * n);
    int flag = convex_hull();
    if (flag == n)
      printf("1\n");
    else
      printf("0\n");
  }
  return 0;
}
