Problem:
Score : 700 points

Problem Statement

You are given N points (x_i,y_i) located on a two-dimensional plane.
Consider a subset S of the N points that forms a convex polygon.
Here, we say a set of points S forms a convex polygon when there exists a convex polygon with a positive area that has the same set of vertices as S. All the interior angles of the polygon must be strictly less than 180°.

For example, in the figure above, {A,C,E} and {B,D,E} form convex polygons; {A,C,D,E}, {A,B,C,E}, {A,B,C}, {D,E} and {} do not.

For a given set S, let n be the number of the points among the N points that are inside the convex hull of S (including the boundary and vertices). Then, we will define the score of S as 2^{n-|S|}.

Compute the scores of all possible sets S that form convex polygons, and find the sum of all those scores.

However, since the sum can be extremely large, print the sum modulo 998244353.

Constraints

1≤N≤200

0≤x_i,y_i<10^4 (1≤i≤N)

If i≠j, x_i≠x_j or y_i≠y_j.

x_i and y_i are integers.

Input

The input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the sum of all the scores modulo 998244353.

Sample Input 1

4
0 0
0 1
1 0
1 1

Sample Output 1

5

We have five possible sets as S, four sets that form triangles and one set that forms a square. Each of them has a score of 2^0=1, so the answer is 5.

Sample Input 2

5
0 0
0 1
0 2
0 3
1 1

Sample Output 2

11

We have three "triangles" with a score of 1 each, two "triangles" with a score of 2 each, and one "triangle" with a score of 4. Thus, the answer is 11.

Sample Input 3

1
3141 2718

Sample Output 3

0

There are no possible set as S, so the answer is 0.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define maxn 205
#define mod 998244353
int x[maxn],y[maxn];
int pow(int a,int n)
{
    if(n==0)return 1;
    int res=pow(a,n/2);
    res=res*res%mod;
    if(n&1)res=res*a%mod;
    return res;
}
int add(int &a,int b)
{
    a+=b;
    if(a>=mod)a%=mod;
}
main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
    int ans=pow(2,n);
    add(ans,mod-1),add(ans,mod-n),add(ans,mod-n*(n-1)/2);
    for(int i=1;i<n;i++)for(int j=i+1;j<=n;j++)
    {
        int cnt=2;
        int a=-(y[j]-y[i]),b=x[j]-x[i],c=x[j]*y[i]-x[i]*y[j];
        for(int k=1;k<=n;k++)
        {
            if(k==i || k==j)continue;
            int s=a*x[k]+b*y[k]-c;
            if(s!=0)continue;
            if(k<j)break;
            cnt++;
        }
        add(ans,mod-pow(2,cnt));
        add(ans,1);
        add(ans,cnt);
        add(ans,cnt*(cnt-1)/2);
    }
    printf("%d\n",ans);
    return 0;
}