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 <algorithm>
#include <complex>
#include <vector>
using namespace std;
const double eps = 1e-10;
#define EQ(a, b) (abs((a) - (b)) < eps)
using P = complex<double>;      // position
struct L : vector<P> {
    L(const P &a = P(), const P &b = P()) {
        emplace_back(a); emplace_back(b);
    }
};
using Polygon = vector<P>;
ostream &operator<<(ostream &os, P p) { return os << real(p) << " " << imag(p); }
double dot(const P &a, const P &b) { return real(conj(a) * b); }
double cross(const P &a, const P &b) { return imag(conj(a) * b); }

enum CCW_RESULT { CCW = +1, CW = -1, BEHIND = +2, FRONT = -2, ON = 0 };
int ccw(P a, P b, P c) {
    b -= a; c -= a;
    if (cross(b, c) > eps) return CCW;      // counter clockwise
    if (cross(b, c) < -eps) return CW;      // clockwise
    if (dot(b, c) < 0) return BEHIND;       // c--a--b on line
    if (norm(b) < norm(c)) return FRONT;    // a--b--c on line | a=b
    return ON;                              // on segment | a=c | b=c
}
#define curr(ps, i) ps[(i) % ps.size()]
#define next(ps, i) ps[(i + 1) % ps.size()]
#define prev(ps, i) ps[(i + ps.size() - 1) % ps.size()]
bool is_convex(const Polygon &ps) {
    for (int i = 0; i < ps.size(); i++) {
        if (ccw(prev(ps, i), curr(ps, i), next(ps, i)) == -1) return false;
    }
    return true;
}

#include <iostream>
#include <iomanip>
int main() {
    cout << fixed << setprecision(1);
    int n; cin >> n;
    Polygon ps;
    for (int i = 0; i < n; i++) {
        int x, y; cin >> x >> y;
        ps.emplace_back(x, y);
    }
    cout << is_convex(ps) << endl;
    return 0;
}
