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/extc++.h>
#include <bits/stdc++.h>

using namespace std;

typedef int64_t ftype;
typedef complex<ftype> point;
#define x real
#define y imag

ftype dot(point a, point b)
{
	return (conj(a) * b).x();
}

ftype cross(point a, point b)
{
	return (conj(a) * b).y();
}

const int mod = 998244353;

void add_mod(int &a, int b)
{
	a += b;
	if(a >= mod)
		a -= mod;
}

signed main()
{
	//freopen("input.txt", "r", stdin);
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	point r[n];
	for(int i = 0; i < n; i++)
	{
		int x, y;
		cin >> x >> y;
		r[i] = {x, y};
	}
	int ans = 0;
	for(int s = 0; s < n; s++)
	{
		point o = r[s];
		vector<point> pts;
		for(int i = 0; i < n; i++)
			if(r[i].y() > o.y() || r[i].y() == o.y() && r[i].x() > o.x())
				pts.push_back(r[i]);
		sort(begin(pts), end(pts), [&](point a, point b)
			{return cross(a - o, b - o) < 0 || cross(a - o, b - o) == 0 && norm(a) < norm(b);});
		int m = pts.size();
		int dp[m][m];
		for(int i = 0; i < m; i++)
			for(int j = 0; j < m; j++)
				dp[i][j] = 0;
		for(int r = 0; r < m; r++)
			for(int l = 0; l < r; l++)
			{
				if(cross(pts[r] - o, pts[l] - o) == 0)
					continue;
				dp[l][r] = 1;
				for(int p = 0; p < l; p++)
					if(cross(pts[r] - pts[l], pts[p] - pts[l]) <= 0)
						add_mod(dp[l][r], dp[l][r]);
				for(int p = 0; p < l; p++)
					if(cross(pts[l] - pts[p], pts[r] - pts[p]) < 0)
						add_mod(dp[l][r], dp[p][l]);
				for(int p = l + 1; p < r; p++)
					if(cross(pts[r] - pts[l], pts[p] - pts[l]) <= 0)
						add_mod(dp[l][r], dp[l][r]);
				add_mod(ans, dp[l][r]);
			}
	}
	cout << ans << endl;
    return 0;
}