#include <bits/stdc++.h>
using namespace std;

const int mo=998244353;
const double pi=acos(-1),eps=1e-10;
int n,x[205],y[205],s[205],ans;
double a[205]; 

int qpow(int a,int b)
{
	int x=a;  a=1;
	while (b)
		{
			if (b&1)  a=1LL*a*x%mo;
			x=1LL*x*x%mo,b>>=1;
		}
	return a;
}

int f(int x){return (qpow(2,x)+mo-(1+x+1LL*x*(x-1)/2)%mo)%mo;}
int dcmp(double x){return (x>eps)-(x<-eps);}

void work()
{
	scanf("%d",&n),ans=f(n);
	for (int i=1; i<=n; i++)  scanf("%d %d",&x[i],&y[i]);
	for (int i=1; i<=n; i++)
		{
			for (int j=1,k=0; j<=n; j++)
				if (i!=j)  a[++k]=atan2(y[i]-y[j],x[i]-x[j]);
			for (int j=1; j<n; j++)  a[j]+=pi*(dcmp(a[j])<=0);
			sort(a+1,a+n);
			for (int l,r=1; r<n; )
				{
					l=r;
					while ((r<n)&&(dcmp(a[r]-a[l])==0))  r++;
					s[r-l+1]++;
				}
		}
	for (int i=1; i<=n; i++)  ans=(ans+mo-1LL*s[i]/i*f(i)%mo)%mo;
	printf("%d",ans);
}

int main()
{
	work();
	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>
using namespace std;

const int mo=998244353;
const double pi=acos(-1),eps=1e-10;
int n,x[205],y[205],s[205],ans;
double a[205]; 

int qpow(int a,int b)
{
	int x=a;  a=1;
	while (b)
		{
			if (b&1)  a=1LL*a*x%mo;
			x=1LL*x*x%mo,b>>=1;
		}
	return a;
}

int f(int x){return (qpow(2,x)+mo-(1+x+1LL*x*(x-1)/2)%mo)%mo;}
int dcmp(double x){return (x>eps)-(x<-eps);}

void work()
{
	scanf("%d",&n),ans=f(n);
	for (int i=1; i<=n; i++)  scanf("%d %d",&x[i],&y[i]);
	for (int i=1; i<=n; i++)
		{
			for (int j=1,k=0; j<=n; j++)
				if (i!=j)  a[++k]=atan2(y[i]-y[j],x[i]-x[j]);
			for (int j=1; j<n; j++)  a[j]+=pi*(dcmp(a[j])<=0);
			sort(a+1,a+n);
			for (int l,r=1; r<n; )
				{
					l=r;
					while ((r<n)&&(dcmp(a[r]-a[l])==0))  r++;
					s[r-l+1]++;
				}
		}
	for (int i=1; i<=n; i++)  ans=(ans+mo-1LL*s[i]/i*f(i)%mo)%mo;
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}

```