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>
using namespace std;
const int limN = 105;
typedef pair<int,int> pii;
#define x first
#define y second

int area(pii a, pii b, pii c) {
    return a.x*b.y + b.x*c.y + c.x*a.y -
        (a.y*b.x + b.y*c.x + c.y*a.x);
}

int sgno(int x) {
    if(x==0)
        return x;
    return x<0? -1 : 1;
}

int N ;
pii ptos[limN];

bool funca(int proh) {
    for(int i=0; i<N; i++) {
        if(sgno(area(ptos[(i+N-1)%N], ptos[i], ptos[(i+1)%N])) == proh)
            return false;
    }
    return true;
}

int main() {
    scanf("%d", &N);
    for(int i=0; i<N; i++)
        scanf("%d%d", &ptos[i].x, &ptos[i].y);
    printf("%d\n", (funca(1) || funca(-1))? 1 : 0);
}

