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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <utility>
#include <queue>
#include <set>
#include <map>
#include <deque>
#include <iomanip>
#include <cstdio>
#include <stack>
#include <numeric>

using namespace std;
typedef  long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define  MP make_pair
#define  PB push_back
#define inf 1000000007
#define mod 998244353
#define rep(i,n) for(int i=0;i<(int)(n);++i)

ll ruijo(ll x,ll n){
	long long p,ans;
	if(n==0)return 1;
	if(n==1)return x;
	if(n%2){
		p = ruijo(x,n/2);
		ans = (x*p)%mod;
		ans = (ans*p)%mod;
		return ans;
	}else{
		p = ruijo(x,n/2);
		ans = (p*p)%mod;
		return ans;
	}
}

int main(){
    int n;
    cin >> n;
    if(n<=2){
        cout << 0 << endl;
        return 0;
    }
    vector<int> x(n),y(n);
    rep(i,n){
        cin >> x[i] >> y[i];
    }
    bool flag[210][210]={};
    vector<int> c;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            vector<int> p;
            if(flag[i][j])continue;
            p.push_back(i);
            p.push_back(j);
            for(int k=j+1;k<n;k++){
                if((x[i]-x[j])*(y[i]-y[k])==(x[i]-x[k])*(y[i]-y[j])){
                    p.push_back(k);
                }
            }
            for(int s=0;s<p.size();s++){
                for(int t = s+1;t<p.size();t++){
                    flag[s][t] = 1;
                }
            }
            c.push_back(p.size());
        }
    }
    ll res = ruijo(2,n);
    res += mod-1-n;
    res %= mod;
    for(int x:c){
        res += 2*mod - (ruijo(2,x)-x-1);
        res %= mod;
    }
    cout << res << endl;
    return 0;
}