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 <stdio.h>
#include <algorithm>
#include <map>
using namespace std;

int N,X[202],Y[202];

int gcd(int a, int b)
{
	return b ? gcd(b,a%b) : a;
}

int main()
{
	scanf ("%d",&N);
	for (int i=0;i<N;i++) scanf ("%d %d",&X[i],&Y[i]);

	long long mod = 998244353;
	long long pw[202] = {1,};
	for (int i=1;i<=N;i++) pw[i] = pw[i-1] * 2 % mod;

	long long ans = (pw[N] - 1 + mod - N) % mod;

	for (int i=0;i<N;i++){
		map<pair<int, int>, int> chk;
		for (int j=i+1;j<N;j++){
			int dx = X[j] - X[i], dy = Y[j] - Y[i];
			int g = gcd(abs(dx),abs(dy));
			dx /= g; dy /= g;
			if (dx < 0) dx = -dx, dy = -dy;
			chk[{dx,dy}]++;
		}

		for (auto &p : chk) ans = (ans + mod - pw[p.second] + 1) % mod;
	}
	
	printf ("%lld\n",ans);
	return 0;
}