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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using uint = unsigned int;
using vi = vector<int>;
using vb = vector<bool>;
using vd = vector<double>;
using vl = vector<ll>;
using vvi = vector<vi>;
using vvb = vector<vb>;
using vvd = vector<vd>;
using vvl = vector<vl>;
template <typename T>
inline string join(const vector<T>& vec, string sep = " ") {
  stringstream ss;
  for (ll i = 0; i < (vec.size()); ++i)
    ss << vec[i] << (i + 1 == vec.size() ? "" : sep);
  return ss.str();
}
template <class T>
using CR = const T&;
using P = complex<ld>;
using G = vector<P>;
int sgn(ld a, ld b = 0) {
  if (a > b + (1e-10)) return 1;
  if (a < b - (1e-10)) return -1;
  return 0;
}
ld dot(P a, P b) { return real(conj(a) * b); }
ld cross(P a, P b) { return imag(conj(a) * b); }
int ccw(P a, P b, P c) {
  b -= a;
  c -= a;
  if (sgn(cross(b, c))) return sgn(cross(b, c));
  if (sgn(dot(b, c)) == -1) return 2;
  if (sgn(norm(b), norm(c)) == -1) return -2;
  return 0;
}
bool is_convex(CR<G> g) {
  for (ll i = 0; i < (g.size()); ++i)
    if (ccw((g[(i + g.size() - 1) % g.size()]), g[i],
            (g[(i + 1) % g.size()])) != 1)
      return false;
  return true;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(30);
  ll n;
  cin >> n;
  G g(n);
  for (ll i = 0; i < (n); ++i) {
    ld x, y;
    cin >> x >> y;
    g[i] = {x, y};
  }
  cout << is_convex(g) << endl;
  return 0;
}
