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>
using namespace std;
const int maxn=2000;
typedef long long ll;
const ll mod=998244353;
ll x[maxn],y[maxn],power[maxn];
/*ll power(ll a,ll b)
{
    ll ret;
    if (b==0) return 1LL;
    ret=pow(a,b/2);
    ret=ret*ret%mod;
    if (b%2) ret=ret*a%mod;
    return ret;
}*/
int main()
{
  //  freopen("text.in","r",stdin);
    ll i,ans,tmp,j,k,n;
    scanf("%lld",&n);
    power[0]=1LL;
    for (i=1;i<=n;i++) {
        power[i]=power[i-1]*2LL%mod;
        scanf("%lld%lld",&x[i],&y[i]);
    }
    ans=((power[n]-n-1)%mod+mod)%mod;
    ans=((ans-(n-1)*n/2)%mod+mod)%mod;
    for (i=1;i<=n;i++) {
        for (j=i+1;j<=n;j++) {
            tmp=0;
            for (k=j+1;k<=n;k++) {
                if ((y[j]-y[i])*(x[k]-x[j])==(y[k]-y[j])*(x[j]-x[i])) tmp++;
                //if ((y[k]-y[i])*(x[j]-x[i])==(x[k]-x[i])*(y[j]-y[i])) tmp++;
            }
            //ans=((ans-/*power(2LL,tmp)*/power[tmp])%mod+mod)%mod;
            ans=((ans-/*power(2LL,tmp)*/power[tmp]+1)%mod+mod)%mod;
        }
    }
    printf("%lld\n",(ans%mod+mod)%mod);
    return 0;
}