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>
#define ll long long
#define inf 0x3f3f3f3f
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
#define pa pair<int,int>
const int N=2e5+10;
const int mod=998244353;
using namespace std;


struct point{ int x,y;}p[N];
inline ll cross(point a,point b,point c){return 1LL*(b.x-a.x)*(c.y-a.y)-1LL*(b.y-a.y)*(c.x-a.x); }
 
int main()
{     
     int n; scanf("%d",&n);
     for(int i=0;i<n;i++)  scanf("%d%d",&p[i].x,&p[i].y);
     int ff=1;
     for(int i=0;i<n;i++)
     {
       int j=(i+1)%n,k=(i+2)%n;
       if(cross(p[i],p[j],p[k])<0)  {ff=0; break;}
     }
     if(ff)puts("1");else puts("0");
return 0;
}
/*

*/
