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;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
const int mod=998244353,MAX=1005,INF=1<<30;
const long double eps=0.000000000003;

ll rui(ll a,ll b){
    ll ans=1;
    while(b>0){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b/=2;
    }
    return ans;
}

int main(){
    
    int N;cin>>N;
    vector<long double> X(N),Y(N);
    for(int i=0;i<N;i++){
        cin>>X[i]>>Y[i];
    }
    
    ll ans=rui(2,N);
    ans+=mod-(N+1);
    ans%=mod;
    
    for(int a=0;a<N;a++){
        for(int b=a+1;b<N;b++){
            ll cnt=0;
            for(int k=0;k<N;k++){
                if(k==a||k==b) continue;
                long double i=atan2(X[b]-X[k],Y[b]-Y[k]),j=atan2(X[a]-X[k],Y[a]-Y[k]);
                if(i>=0&&j<=0) swap(i,j);
                if(i*j<0||(i==0&&j<0)||(i<0&&j==0)){
                    if(abs(abs(i)-j)<=eps) cnt++;
                }
            }
            ans+=mod-rui(2,cnt);
            ans%=mod;
        }
    }
    
    cout<<ans<<endl;
}