#include <bits/stdc++.h>
#define ll long long
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i <= (n); ++i)
#define ms(x, v) memset(x, v, sizeof(x))
using namespace std;
const int N = 205, MOD = 998244353;
pair<int, int> s[N];
int operator ^ (pair<int, int> a, pair<int, int> b)
{
	return a.first * b.second - a.second * b.first;
}
pair<int, int> operator - (pair<int, int> a, pair<int, int> b)
{
	return make_pair(a.first - b.first, a.second - b.second);
}
void add(int &a, int b) { a = (a + b) % MOD; }
int pw[N];
int main()
{
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(0), cin.tie(0);
	int n;
	cin >> n;
	rep1(i, n) cin >> s[i].first >> s[i].second;
	pw[0] = 1;
	rep1(i, n) pw[i] = pw[i - 1] * 2 % MOD;
	int ans = pw[n];
	add(ans, MOD - 1);
	add(ans, MOD - n);
	rep1(i, n)
	{
		rep1(j, n)
		{
			if (s[i] < s[j])
			{
				int cnt = 0;
				rep1(k, n)
				{
					if (s[i] < s[k] && s[j] < s[k] && ((s[i] - s[k]) ^ (s[j] - s[k])) == 0) cnt++;
				}
				add(ans, MOD - pw[cnt]);
			}
		}
	}
	cout << ans << endl;
	return 0;
}
### Prompt

Generate a Cpp solution to the following problem:

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°.

cddb0c267926c2add885ca153c47ad8a.png

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.

Examples

Input

4
0 0
0 1
1 0
1 1


Output

5


Input

5
0 0
0 1
0 2
0 3
1 1


Output

11


Input

1
3141 2718


Output

0

### Response

```cpp
#include <bits/stdc++.h>
#define ll long long
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i <= (n); ++i)
#define ms(x, v) memset(x, v, sizeof(x))
using namespace std;
const int N = 205, MOD = 998244353;
pair<int, int> s[N];
int operator ^ (pair<int, int> a, pair<int, int> b)
{
	return a.first * b.second - a.second * b.first;
}
pair<int, int> operator - (pair<int, int> a, pair<int, int> b)
{
	return make_pair(a.first - b.first, a.second - b.second);
}
void add(int &a, int b) { a = (a + b) % MOD; }
int pw[N];
int main()
{
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
#endif
	ios::sync_with_stdio(0), cin.tie(0);
	int n;
	cin >> n;
	rep1(i, n) cin >> s[i].first >> s[i].second;
	pw[0] = 1;
	rep1(i, n) pw[i] = pw[i - 1] * 2 % MOD;
	int ans = pw[n];
	add(ans, MOD - 1);
	add(ans, MOD - n);
	rep1(i, n)
	{
		rep1(j, n)
		{
			if (s[i] < s[j])
			{
				int cnt = 0;
				rep1(k, n)
				{
					if (s[i] < s[k] && s[j] < s[k] && ((s[i] - s[k]) ^ (s[j] - s[k])) == 0) cnt++;
				}
				add(ans, MOD - pw[cnt]);
			}
		}
	}
	cout << ans << endl;
	return 0;
}
```