#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define Vector Point
const int MOD=998244353;
const int MAXN=200;
void M(int& a){((a%=MOD)+=MOD)%=MOD;}
struct Point{int x,y;};
Vector operator-(const Point& a,const Point& b){return (Vector){a.x-b.x,a.y-b.y};}
int cross(const Vector& a,const Vector& b){return a.x*b.y-a.y*b.x;}
int n,ans;
Point p[MAXN+10];
int quickpow(int x,int y)
{
	int res=1;
	for(;y;y>>=1,x=(ll)x*x%MOD)if(y&1)res=(ll)res*x%MOD;
	return res;
}
int main()
{
	scanf("%d",&n);M(ans=quickpow(2,n)-n-1);
	for(int i=1;i<=n;++i)scanf("%d%d",&p[i].x,&p[i].y);
	Point a,b;int cnt;bool flag;
	for(int i=1;i<=n;++i)
	{
		for(int j=i+1;j<=n;++j)
		{
			flag=true;a=p[i];b=p[j];
			for(int k=1;k<j;++k)if(k!=i&&cross(a-p[k],b-p[k])==0){flag=false;break;}
			if(!flag)continue;
			cnt=2;
			for(int k=j+1;k<=n;++k)if(cross(a-p[k],b-p[k])==0)++cnt;
			M(ans-=quickpow(2,cnt)-cnt-1);
		}
	}
	printf("%d",ans);
	return 0;
}
### Prompt

Create a solution in Cpp for 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<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define ll long long
#define Vector Point
const int MOD=998244353;
const int MAXN=200;
void M(int& a){((a%=MOD)+=MOD)%=MOD;}
struct Point{int x,y;};
Vector operator-(const Point& a,const Point& b){return (Vector){a.x-b.x,a.y-b.y};}
int cross(const Vector& a,const Vector& b){return a.x*b.y-a.y*b.x;}
int n,ans;
Point p[MAXN+10];
int quickpow(int x,int y)
{
	int res=1;
	for(;y;y>>=1,x=(ll)x*x%MOD)if(y&1)res=(ll)res*x%MOD;
	return res;
}
int main()
{
	scanf("%d",&n);M(ans=quickpow(2,n)-n-1);
	for(int i=1;i<=n;++i)scanf("%d%d",&p[i].x,&p[i].y);
	Point a,b;int cnt;bool flag;
	for(int i=1;i<=n;++i)
	{
		for(int j=i+1;j<=n;++j)
		{
			flag=true;a=p[i];b=p[j];
			for(int k=1;k<j;++k)if(k!=i&&cross(a-p[k],b-p[k])==0){flag=false;break;}
			if(!flag)continue;
			cnt=2;
			for(int k=j+1;k<=n;++k)if(cross(a-p[k],b-p[k])==0)++cnt;
			M(ans-=quickpow(2,cnt)-cnt-1);
		}
	}
	printf("%d",ans);
	return 0;
}
```