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>
typedef struct {
  double x, y;
} PP;
typedef struct {
  PP s, e;
} SEG, LINE;
int dcmp(double x) {
  if (fabs(x) < 1e-8) return 0;
  return x <= 0 ? -1 : 1;
}
PP vset(double x, double y) {
  PP r;
  r.x = x, r.y = y;
  return r;
}
PP vadd(PP p1, PP p2) {
  PP r;
  r.x = p1.x + p2.x, r.y = p1.y + p2.y;
  return r;
}
PP vsub(PP p1, PP p2) {
  PP r;
  r.x = p1.x - p2.x, r.y = p1.y - p2.y;
  return r;
}
PP vsmul(PP p, double k) {
  PP r;
  r.x = p.x * k, r.y = p.y * k;
  return r;
}
PP vmul(PP p1, PP p2) {
  PP r;
  r.x = p1.x * p2.x - p1.y * p2.y, r.y = p1.x * p2.y + p1.y * p2.x;
  return r;
}
double vabs(PP a) { return hypot(a.x, a.y); }
double dist(PP p1, PP p2) { return hypot(p1.x - p2.x, p1.y - p2.y); };
double cross(PP a, PP b) { return a.x * b.y - a.y * b.x; }
double dot(PP a, PP b) { return a.x * b.x + a.y * b.y; }
double norm(PP a) { return a.x * a.x + a.y * a.y; }
int in() {
  int n = 0, c = getchar_unlocked();
  if (c == '-') {
    c = getchar_unlocked();
    do n = 10 * n + (c & 0xf), c = getchar_unlocked();
    while (c >= '0');
    return -n;
  }
  do n = 10 * n + (c & 0xf), c = getchar_unlocked();
  while (c >= '0');
  return n;
}
int ccw(PP p0, PP p1, PP p2) {
  PP a, b;
  double t;
  a = vsub(p1, p0), b = vsub(p2, p0), t = cross(a, b);
  if (t > 1e-8) return 1;
  if (t < -1e-8) return -1;
  if (dot(a, b) < -1e-8) return 2;
  if (norm(a) < norm(b)) return -2;
  return 0;
}
int isConvex(int n, PP *p) {
  int i, t;
  p[n] = p[0], p[n + 1] = p[1];
  for (i = 0; i < n; i++) {
    t = ccw(p[i], p[i + 1], p[i + 2]);
    if (t != 1 && t != -2) return 0;
  }
  return 1;
}
PP p[110];
int main() {
  int n, i;
  n = in();
  for (i = 0; i < n; i++) p[i].x = in(), p[i].y = in();
  putchar('0' + isConvex(n, p)), putchar('\n');
  return 0;
}
