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 <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <cstdlib>
#include <queue>
#include <vector>
#include <set>
#include <stdio.h>
#include <cstdio>
#include <cmath>
#include <complex>
using namespace std;
typedef complex<double> xy_t;
xy_t P[30];
int N;
double x,y;
int aa=1;
int main() {
cin >> N;
double s =0.0;
for(int i = 0;i < N;i++) {
    cin >> x >> y;
    P[i] = xy_t(x,y);
}
for ( int i = 0; i < N;i++ ){
    xy_t a = P[(i+1)%N] - P[i], b = P[(i+2)%N]-P[(i+1)%N];
    if((a.real()*b.imag()-a.imag()*b.real())<0){
    	aa=0; break;
    }
}
printf("%d\n", aa);
}