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: Accepted)

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

using namespace std;

typedef long long ll;
#define REP(i,n) for(int i=0,_n=(int)(n);i<_n;++i)
#define ALL(v) (v).begin(),(v).end()
template<class T1,class T2>ostream& operator<<(ostream& os,const pair<T1,T2>&a){return os<<"("<<a.first<<","<<a.second<< ")";}
template<class T>void pv(T a,T b){for(T i=a;i!=b;++i)cout<<(*i)<<" ";cout<<endl;}
template<class T>bool chmin(T&a,const T&b){return a>b?(a=b,1):0;}
template<class T>bool chmax(T&a,const T&b){return a<b?(a=b,1):0;}

const ll MOD = 998244353;

int nextInt() { int x; scanf("%d", &x); return x;}

int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a%b);
}

const int MAX_N = 210;
int X[MAX_N];
int Y[MAX_N];

// (x1, y1) -- (x2, y2) を通る直線の方程式 ax + by + c == 0 の {a,b,c} を返す
vector<int> line(int x1, int y1, int x2, int y2) {
  if (x1 == x2) {
    return {1, 0, -x1};
  }
  if (y1 == y2) {
    return {0, 1, -y1};
  }
  int a = (y2 - y1);
  int b = (x1 - x2);
  int c = y1*x2 - x1*y2;
  if (a < 0) {
    a *= -1;
    b *= -1;
    c *= -1;
  }
  int g = gcd(a, gcd(abs(b), abs(c)));
  a /= g;
  b /= g;
  c /= g;
  return {a, b, c};
}

ll two[100000];

int main2() {
  two[0] = 1;
  for (int i = 1; i < 100000; i++) two[i] = 2 * two[i-1] % MOD;

  int N = nextInt();
  REP(i, N) {
    X[i] = nextInt();
    Y[i] = nextInt();
  }
  map<vector<int>, int> mp;
  REP(i, N) REP(j, i) {
    vector<int> l = line(X[i], Y[i], X[j], Y[j]);
    ++mp[l];
  }

  ll ans = two[N] - 1 - N - N*(N-1)/2;
  for(auto it = mp.begin(); it != mp.end(); ++it) {
    int m = (1 + (int)sqrt(1.5 + 8 * (it->second))) / 2;
    ans -= two[m] - 1 - m - (m)*(m-1)/2;
    ans = (ans % MOD + MOD) % MOD;
  }
  ans %= MOD;
  cout << ans << endl;
  return 0;
}

int main() {
  for (;!cin.eof();cin>>ws) main2();
  return 0;
}