#include <iostream>
#include <cstdio>

using namespace std;

const int P=998244353;
const int N=205;

int x[N],y[N];
int C[N][N];
int n,ans,pw;

int main()
{
	//freopen("cvxscr.in","r",stdin),freopen("cvxscr.out","w",stdout);
	scanf("%d",&n),pw=1;
	for (int i=1;i<=n;++i) scanf("%d%d",&x[i],&y[i]),(pw<<=1)%=P;
	C[0][0]=1;
	for (int i=1;i<=n;++i)
	{
		C[i][0]=1;
		for (int j=1;j<=i;++j) C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;
	}
	ans=n+1;
	for (int i=1;i<n;++i)
		for (int j=i+1;j<=n;++j)
		{
			int cnt=2,vx=x[i]-x[j],vy=y[i]-y[j];
			for (int k=1,vx_,vy_;~cnt&&k<i;++k)
			{
				vx_=x[i]-x[k],vy_=y[i]-y[k];
				if (!(vx*vy_-vy*vx_)) cnt=-1;
			}
			for (int k=i+1,vx_,vy_;~cnt&&k<j;++k)
			{
				vx_=x[i]-x[k],vy_=y[i]-y[k];
				if (!(vx*vy_-vy*vx_)) cnt=-1;
			}
			if (!~cnt) continue;
			for (int k=j+1,vx_,vy_;k<=n;++k) vx_=x[i]-x[k],vy_=y[i]-y[k],cnt+=!(vx*vy_-vy*vx_);
			for (int k=2;k<=cnt;++k) (ans+=C[cnt][k])%=P;
		}
	ans=(P-ans+pw)%P,printf("%d\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
### Prompt

Please formulate 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 <iostream>
#include <cstdio>

using namespace std;

const int P=998244353;
const int N=205;

int x[N],y[N];
int C[N][N];
int n,ans,pw;

int main()
{
	//freopen("cvxscr.in","r",stdin),freopen("cvxscr.out","w",stdout);
	scanf("%d",&n),pw=1;
	for (int i=1;i<=n;++i) scanf("%d%d",&x[i],&y[i]),(pw<<=1)%=P;
	C[0][0]=1;
	for (int i=1;i<=n;++i)
	{
		C[i][0]=1;
		for (int j=1;j<=i;++j) C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;
	}
	ans=n+1;
	for (int i=1;i<n;++i)
		for (int j=i+1;j<=n;++j)
		{
			int cnt=2,vx=x[i]-x[j],vy=y[i]-y[j];
			for (int k=1,vx_,vy_;~cnt&&k<i;++k)
			{
				vx_=x[i]-x[k],vy_=y[i]-y[k];
				if (!(vx*vy_-vy*vx_)) cnt=-1;
			}
			for (int k=i+1,vx_,vy_;~cnt&&k<j;++k)
			{
				vx_=x[i]-x[k],vy_=y[i]-y[k];
				if (!(vx*vy_-vy*vx_)) cnt=-1;
			}
			if (!~cnt) continue;
			for (int k=j+1,vx_,vy_;k<=n;++k) vx_=x[i]-x[k],vy_=y[i]-y[k],cnt+=!(vx*vy_-vy*vx_);
			for (int k=2;k<=cnt;++k) (ans+=C[cnt][k])%=P;
		}
	ans=(P-ans+pw)%P,printf("%d\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
```