#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int N=205;
const int MOD=998244353;
int n;
int x[N],y[N];
int Pow[N];
int mul (int a,int b,int c)
{
	int x_1=x[a]-x[c],y_1=y[a]-y[c];
	int x_2=x[b]-x[c],y_2=y[b]-y[c];
	return x_1*y_2-x_2*y_1;
}
bool check (int x,int y)
{
	for (int u=1;u<y;u++)
	{
		if (u==x) continue;
		if (mul(x,y,u)==0)	
		{
		//	printf("NO:%d %d %d\n",x,y,u);
			return false;
		}
	}
	return true;
}
int dec (int x,int y)	{x=x-y;return x<0?x+MOD:x;}
int main()
{
	scanf("%d",&n);
	Pow[0]=1;for (int u=1;u<=n;u++) Pow[u]=Pow[u-1]*2%MOD;
	for (int u=1;u<=n;u++) scanf("%d%d",&x[u],&y[u]);
	int ans=dec(Pow[n],n+1);
	for (int u=1;u<=n;u++)
		for (int i=u+1;i<=n;i++)
			if (check(u,i))//这条线有没有枚举过
			{
			//	printf("YES:%d %d\n",u,i);
				int cnt=2;
				for (int j=i+1;j<=n;j++) cnt=cnt+(mul(u,i,j)==0);
				ans=dec(ans,dec(Pow[cnt],cnt+1));
			}
	printf("%d\n",ans);
	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<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int N=205;
const int MOD=998244353;
int n;
int x[N],y[N];
int Pow[N];
int mul (int a,int b,int c)
{
	int x_1=x[a]-x[c],y_1=y[a]-y[c];
	int x_2=x[b]-x[c],y_2=y[b]-y[c];
	return x_1*y_2-x_2*y_1;
}
bool check (int x,int y)
{
	for (int u=1;u<y;u++)
	{
		if (u==x) continue;
		if (mul(x,y,u)==0)	
		{
		//	printf("NO:%d %d %d\n",x,y,u);
			return false;
		}
	}
	return true;
}
int dec (int x,int y)	{x=x-y;return x<0?x+MOD:x;}
int main()
{
	scanf("%d",&n);
	Pow[0]=1;for (int u=1;u<=n;u++) Pow[u]=Pow[u-1]*2%MOD;
	for (int u=1;u<=n;u++) scanf("%d%d",&x[u],&y[u]);
	int ans=dec(Pow[n],n+1);
	for (int u=1;u<=n;u++)
		for (int i=u+1;i<=n;i++)
			if (check(u,i))//这条线有没有枚举过
			{
			//	printf("YES:%d %d\n",u,i);
				int cnt=2;
				for (int j=i+1;j<=n;j++) cnt=cnt+(mul(u,i,j)==0);
				ans=dec(ans,dec(Pow[cnt],cnt+1));
			}
	printf("%d\n",ans);
	return 0;
}
```