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>
#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define mkp make_pair
typedef long long ll;
using namespace std;

int N, x[200], y[200];
ll res, pw2[200];

ll pw(ll x, int y) {
	ll a = 1;
	while (y) {
		if (y & 1) {
			a = a*x%mod;
		}
		x = x*x%mod;
		y >>= 1;
	}
	return a;
}

ll modinv(ll x) {
	return pw(x, mod - 2);
}

int main(){
	cin >> N;
	ll t = N*(N - 1), m = modinv(2);
	for (int i = 3; i <= N; ++i) {
		t = (t*(N + 1 - i)) % mod;
		m = (m*modinv(i)) % mod;
		res = (res + t*m) % mod;
	}
	pw2[0] = 1;
	for (int i = 1; i < 200; ++i) {
		pw2[i] = pw2[i - 1] * 2 % mod;
	}
	for (int i = 0; i < N; ++i) {
		cin >> x[i] >> y[i];
	}
	for (int i = 0; i < N; ++i) {
		for (int j = i + 1; j < N; ++j) {
			int t = 0;
			for (int k = j + 1; k < N; ++k) {
				if ((x[i] - x[j])*(y[j] - y[k]) == (x[j] - x[k])*(y[i] - y[j])) {
					res = (res + mod - pw2[t]) % mod;
					++t;
				}
			}
		}
	}
	cout << res << endl;
	return 0;
}