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

const int mo=998244353;

int n,Ans;

struct Point{
	int x,y;
	Point() {}
	Point(int x,int y):x(x),y(y) {}
	Point operator + (const Point &p) {return Point(x+p.x,y+p.y);}
}P[210];

int power(int x,int k)
{
	int res=1;
	while (k)
	{
		if (k&1) res=(LL)res*x%mo;
		x=(LL)x*x%mo;
		k>>=1;
	}
	return res;
}

int main()
{
#ifdef h10
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int i,j,k;
	scanf("%d",&n);
	for (i=1;i<=n;i++)
		scanf("%d%d",&P[i].x,&P[i].y);
	Ans=power(2,n);
	for (i=1;i<=n;i++)
	for (j=i+1;j<=n;j++)
	{
		int cnt=0;
		for (k=j+1;k<=n;k++)
			cnt+=(P[i].x-P[j].x)*(P[i].y-P[k].y)==(P[i].x-P[k].x)*(P[i].y-P[j].y);
		(Ans+=mo-power(2,cnt))%=mo;
	}
	(Ans+=mo-n-1)%=mo;
	printf("%d\n",Ans);
}

### 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<bits/stdc++.h>
using namespace std;
#define LL long long

const int mo=998244353;

int n,Ans;

struct Point{
	int x,y;
	Point() {}
	Point(int x,int y):x(x),y(y) {}
	Point operator + (const Point &p) {return Point(x+p.x,y+p.y);}
}P[210];

int power(int x,int k)
{
	int res=1;
	while (k)
	{
		if (k&1) res=(LL)res*x%mo;
		x=(LL)x*x%mo;
		k>>=1;
	}
	return res;
}

int main()
{
#ifdef h10
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int i,j,k;
	scanf("%d",&n);
	for (i=1;i<=n;i++)
		scanf("%d%d",&P[i].x,&P[i].y);
	Ans=power(2,n);
	for (i=1;i<=n;i++)
	for (j=i+1;j<=n;j++)
	{
		int cnt=0;
		for (k=j+1;k<=n;k++)
			cnt+=(P[i].x-P[j].x)*(P[i].y-P[k].y)==(P[i].x-P[k].x)*(P[i].y-P[j].y);
		(Ans+=mo-power(2,cnt))%=mo;
	}
	(Ans+=mo-n-1)%=mo;
	printf("%d\n",Ans);
}

```