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<functional>
#include<algorithm>
#include<iostream>
#include<utility>
#include<climits>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<cctype>
#include<cmath>
#include<stack>
#include<queue>
#include<list>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const int inf88 = 1482184792;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef pair<int, int> P;
#define N 1010
#define M 1000100

int sgn(double x){
    if(fabs(x) < eps)
        return 0;
    if(x < 0)
        return -1;
    return 1;
}

int n;

struct Point{
    double x, y;
    Point(){};
    Point(double _x, double _y){
        x = _x;
        y = _y;
    }
    bool operator < (const Point &A) const{
        if(x == A.x)
            return y < A.y;
        return x < A.x;
    }
    bool operator == (Point A) const{
        return sgn(x-A.x) == 0 && sgn(y-A.y) == 0;
    }
    Point operator - (const Point &A) const{
        return Point(x-A.x, y-A.y);
    }
    double operator * (const Point &A) const{
        return x*A.x + y*A.y;
    }
    double operator ^ (const Point &A) const{
        return x*A.y - y*A.x;
    }
    double distance(Point A){
        return sqrt((x-A.x)*(x-A.x) + (y-A.y)*(y-A.y));
    }
}p[N], sta[N];

int cnt;

double xmul(Point p0, Point p1, Point p2){
    return (p1-p0)^(p2-p0);
}

bool convex_hull(){
    map<int, int> dir;
    for(int i = 0; i < n; i++){
        int tmp = sgn(xmul(p[i+1], p[i], p[i+2]));
        dir[tmp]++;
        if(dir[1] && dir[-1])
            return false;
    }
    return true;
}

int main(){
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++)
            scanf("%lf%lf", &p[i].x, &p[i].y);
        p[0].x = p[n].x; p[0].y = p[n].y;
        p[n+1].x = p[1].x; p[n+1].y = p[1].y;
        if(convex_hull())
            printf("1\n");
        else
            printf("0\n");
    }
    return 0;
}

