#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int m[200][200];
int x[200], y[200];
typedef long long ll;
ll p2[1000];
ll mod = 998244353;
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)scanf("%d%d", &x[i], &y[i]);
	p2[0] = 1;
	for (int i = 1; i < 1000; i++)p2[i] = p2[i - 1] * 2 % mod;
	ll s = p2[num];
	for (int i = 0; i < num; i++)
	{
		for (int j = i + 1; j < num; j++)
		{
			if (m[i][j])continue;
			vector<int>v;
			for (int k = 0; k < num; k++)
			{
				if ((x[k] - x[i])*(y[j] - y[i]) == (y[k] - y[i])*(x[j] - x[i]))v.push_back(k);
			}
			s = (s + mod - p2[v.size()] + v.size() + 1) % mod;
			for (int k = 0; k < v.size(); k++)for (int l = 0; l < v.size(); l++)m[v[k]][v[l]] = 1;
		}
	}
	s = (s + mod - num - 1)%mod;
	printf("%lld\n", s);
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int m[200][200];
int x[200], y[200];
typedef long long ll;
ll p2[1000];
ll mod = 998244353;
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i < num; i++)scanf("%d%d", &x[i], &y[i]);
	p2[0] = 1;
	for (int i = 1; i < 1000; i++)p2[i] = p2[i - 1] * 2 % mod;
	ll s = p2[num];
	for (int i = 0; i < num; i++)
	{
		for (int j = i + 1; j < num; j++)
		{
			if (m[i][j])continue;
			vector<int>v;
			for (int k = 0; k < num; k++)
			{
				if ((x[k] - x[i])*(y[j] - y[i]) == (y[k] - y[i])*(x[j] - x[i]))v.push_back(k);
			}
			s = (s + mod - p2[v.size()] + v.size() + 1) % mod;
			for (int k = 0; k < v.size(); k++)for (int l = 0; l < v.size(); l++)m[v[k]][v[l]] = 1;
		}
	}
	s = (s + mod - num - 1)%mod;
	printf("%lld\n", s);
}
```