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 <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

#define mp make_pair
#define pb push_back
#define ll long long

#define maxN 211
#define mod 998244353

int n, i, j, k;
pair<int, int> P[maxN];
vector< pair<int, int> > ord;

ll pows[maxN];
ll ans, dx, dy, d;

ll gcd(ll a, ll b) {
    if (a < 0) a *= -1;
    if (b < 0) b *= -1;
    if (a < b) swap(a, b);

    while (b > 0) {
        a %= b;
        swap(a, b);
    }

    return a;
}

int main()
{
  //  freopen("test.in","r",stdin);

    cin >> n;
    for (i = 1; i <= n; i++)
        cin >> P[i].first >> P[i].second;

    pows[0] = 1;
    for (i = 1; i <= n; i++)
        pows[i] = (pows[i - 1] * 2) % mod;

    ans = pows[n];

    for (i = 1; i <= n; i++) {
        ord.clear();
        for (j = i + 1; j <= n; j++) {
            dx = P[i].first - P[j].first;
            dy = P[i].second - P[j].second;

            if (dx < 0) {
                dx *= -1;
                dy *= -1;
            }

            if (dx == 0 && dy < 0)
                dy *= -1;

            d = gcd(dx, dy);
            dx /= d;
            dy /= d;
            ord.pb(mp(dx, dy));
        }

        sort(ord.begin(), ord.end());
        for (j = 0; j < ord.size(); j = k) {
            for (k = j; k < ord.size(); k++)
                if (ord[j] != ord[k])
                    break;

            ans += mod - pows[k - j] + 1;
            ans %= mod;
        }

        ans += mod - 1;
        ans %= mod;
    }

    ans += mod - 1;
    ans %= mod;

    cout << ans;

    return 0;
}