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++ (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int N = 211;
const int mod = 998244353;
int n;
LL base[N];
struct ddd{
    int x, y;
}nd[N];
LL sq(LL x){return x * x;}
bool check(LL x1, LL y1, LL x2, LL y2){
    LL a = x1 * x2 + y1 * y2;
    LL b = (sq(x1) + sq(y1)) * (sq(x2) + sq(y2));
    return a > 0 && b == sq(a);
}
int main(){
    cin>>n;
    for(int i = 1;i <= n; i++){
        scanf("%d%d", &nd[i].x, &nd[i].y);
    }
    base[0] = 1;
    for(int i = 1;i <= 200; i++)base[i] = base[i-1] * 2 % mod;
    LL ans = (base[n] - 1 - n + mod) % mod;
    for(int i = 2;i <= n; i++){
        for(int j = 1;j < i; j++){
            int tot = 0;
            for(int k = 1;k <= n; k++){
                if(i == k || j == k)continue;
                if(check(nd[i].x - nd[k].x, nd[i].y - nd[k].y, nd[k].x - nd[j].x, nd[k].y - nd[j].y))tot++;
            }
            ans = (ans + mod - base[tot]) % mod;
            //printf("i=%d j=%d tot=%d\n", i, j, tot);
        }
    }
    cout<<ans<<endl;
    return 0;
}