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>

using namespace std;
using int64 = long long;

constexpr int DEBUG = 0;

struct Int2D {
  const int64 x, y;
  Int2D(int64 x, int64 y) : x(x), y(y) {}
  Int2D operator-(Int2D o) const {
    return Int2D(x - o.x, y - o.y);
  }
  int64 CrossProduct(Int2D o) const {
    return x * o.y - y * o.x;
  }
};

int64 CrossProduct(Int2D p1, Int2D p2) {
  return p1.x * p2.y - p1.y * p2.x;
}

bool Colinear(Int2D p1, Int2D p2, Int2D p3) {
  Int2D v12 = p2 - p1;
  Int2D v13 = p3 - p1;
  return CrossProduct(v12, v13) == 0;
}

constexpr int64 P = 998244353;

struct FiniteField {
 private:
  int64 x;
 public:
  FiniteField(int64 x) : x(x) {
    if (x > P || x < 0) { cerr << "Invalied FiniteField!" << endl; exit(1); }
  }
  FiniteField() : x(0) {}
  int64 Value() const { return x; }
  inline FiniteField operator+(FiniteField o) const {
    FiniteField r(*this); r += o; return r;
  }
  inline FiniteField operator-(FiniteField o) const {
    FiniteField r(*this); r -= o; return r;
  }
  inline FiniteField operator* (FiniteField o) const {
    FiniteField r(*this); r *= o; return r;
  }
  inline FiniteField operator/ (FiniteField o) const {
    FiniteField r(*this); r /= o; return r;
  }
  inline void operator+= (FiniteField o) { x = (x + o.x) % P; }
  inline void operator-= (FiniteField o) { x = (x + P - o.x) % P; }
  inline void operator*= (FiniteField o) { x = (x * o.x) % P; }
  void operator/=(FiniteField o) {
    int64 p = P - 2; while (p) { if (p % 2) { *this *= o; } o *= o; p /= 2; }
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  vector<Int2D> ps;
  for (int i = 0; i < n; i++) {
    int64 x, y;
    cin >> x >> y;
    ps.emplace_back(x, y);
  }

  FiniteField r = 0;
  r += 1;
  r += n;

  set<pair<int, int>> visited;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (visited.find(make_pair(i, j)) != visited.end()) {
        continue;
      }

      vector<int> vs({i, j});
      for (int k = j + 1; k < n; k++) {
        if (Colinear(ps[i], ps[j], ps[k])) {
          vs.push_back(k);
        }
      }

      int m = vs.size();
      FiniteField s = 1;
      for (int k = 0; k < m; k++) s *= 2;
      s -= 1;
      s -= m;

      r += s;

      for (int i1 = 0; i1 < vs.size(); i1++) {
        for (int i2 = i1 + 1; i2 < vs.size(); i2++) {
          visited.insert(make_pair(i1, i2));
        }
      }
    }
  }

  FiniteField ans = 1;
  for (int i = 0; i < n; i++) ans *= 2;
  ans -= r;
  // cout << r.Value() << endl;
  cout << ans.Value() << endl;
}