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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define re(i,a,b) for(int i=(a);i<(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)
#define run(v) for(int k=head[v];k;k=e[k].ne)
#define v e[k].t
#define clr(a) memset(a,0,sizeof(a));
#define il inline
#define sz(a) ((int)a.size())
#define all(a) a.begin(),a.end()
#define mp make_pair
#define pb push_back 
#define w1 first
#define w2 second
#define adm(a,b,c) {a=a+b;if(a>=c)a-=c;else if(a<0)a+=c;}
typedef pair<int,int> pa;
typedef long long ll;typedef long double ld;typedef unsigned long long ull;
const int N=1e5+5,M=5e7+5,INF=1e9,mod=998244353;
const ll linf=1e18;const double eps=1e-8,pi=acos(-1);
il int gmin(int &a,int b){if(a>b)a=b;}il ll gmin(ll &a,ll b){if(a>b)a=b;}il int gmax(int &a,int b){if(a<b)a=b;}il ll gmax(ll &a,ll b){if(a<b)a=b;}
il void read(ll&x){ll f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il ll read(ll&x,ll&y){read(x);read(y);}
il void read(int&x){int f=1,t=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){t=t*10+ch-'0';ch=getchar();}x=t*f;}il int read(int&x,int&y){read(x);read(y);}
il void read(int&a,int&b,int&c){read(a);read(b);read(c);}il void read(ll&a,ll&b,ll&c){read(a);read(b);read(c);}
il int read(){int x;read(x);return x;}
il ll qpow(ll a,ll b,ll p){ll ret=1;for(;b;b>>=1,a=a*a%p)if(b&1)ret=ret*a%p;return ret;}il ll qpow(ll a,ll b){ll ret=1;for(;b;b>>=1,a=a*a%mod)if(b&1)ret=ret*a%mod;return ret;}
il ll qmul(ll a,ll b,ll p){ll ret=0;for(;b;b>>=1,a=(a<<1)%p)if(b&1)adm(ret,a,p);return ret;}il ll qmul(ll a,ll b){ll ret=0;for(;b;b>>=1,a=(a<<1)%mod)if(b&1)adm(ret,a,mod);return ret;}
il void judge(){
	freopen("data.in","r",stdin);
	freopen("data.out","w",stdout);}
il void gen(){freopen("data.in","w",stdout);}
int n;
struct P{
	int x,y;
	il P(){}
	il P(int x,int y):x(x),y(y){}
	il bool operator<(const P&o)const{if(x==o.x)return y<o.y;return x<o.x;}
	il int operator*(const P&o)const{return x*o.y-y*o.x;}
	il P operator-(const P&o)const{return P(x-o.x,y-o.y);}
}p[N];
ll pw[N];
il int bad(int i,int j,int k){
	if((p[j]-p[i])*(p[k]-p[i])==0)return 1;
	return 0;
}
int main(){
	read(n);rep(i,1,n)read(p[i].x,p[i].y);
	pw[0]=1;rep(i,1,n)pw[i]=pw[i-1]*2%mod;
	sort(p+1,p+1+n);
	ll qaq=0;
	rep(i,1,n)rep(j,i+1,n){
		int ok=1;
		rep(k,1,j)
			if(k!=i&&k!=j)
				if(bad(i,j,k))ok=0;
		if(ok){
			int res=2;
			rep(k,j+1,n)res+=bad(i,j,k);
			qaq+=pw[res]-res-1;
		}
	}
	qaq=pw[n]-qaq-n-1;
	cout<<qaq<<endl;
}