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>
#define MOD 998244353LL
#define EPS 1e-9
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

struct data{
	int x,y;
	data(){}
	data(int xx,int yy){
		x=xx;
		y=yy;
	}
	bool operator<(const data &d)const{
		return atan2(y,x)<atan2(d.y,d.x);
	}
};

int n;
data d[205];
ll two[205];
bool flag[205][205];
ll res=0;

bool check(int f,int t,int vx,int vy){
	if(vx==0){
		return d[f].x==d[t].x;
	}
	if(vy==0){
		return d[f].y==d[t].y;
	}
	int sax=d[t].x-d[f].x;
	int say=d[t].y-d[f].y;
	return (sax*vy-say*vx)==0;
}

int main(void){
	scanf("%d",&n);
	two[0]=1;
	for(int i=1;i<=n;i++){
		two[i]=two[i-1]*2LL%MOD;
	}
	for(int i=0;i<n;i++){
		scanf("%d%d",&d[i].x,&d[i].y);
	}
	ll res=two[n];
	for(int i=0;i<n;i++){
		for(int j=i+1;j<n;j++){
			if(flag[i][j])continue;
			int vx=d[i].x-d[j].x;
			int vy=d[i].y-d[j].y;
			vector<int> vec;
			int cnt=0;
			for(int k=0;k<n;k++){
				if(check(j,k,vx,vy)){
					vec.push_back(k);
					cnt++;
				}
			}
			res-=two[cnt];
			res+=MOD;
			res%=MOD;
			res+=cnt+1;
			res%=MOD;
			for(int k=0;k<vec.size();k++){
				for(int l=0;l<vec.size();l++){
					flag[k][l]=true;
				}
			}
		}
	}
	res-=n+1;
	res+=MOD;
	res%=MOD;
	printf("%lld\n",res);
	return 0;
}