#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=201;
typedef long long LL;
const LL mod=998244353;

int n,x[maxn],y[maxn];
LL ans;
bool f;

LL pw(int p)
{
	if (p==1) return 2ll;
	LL ans=pw(p/2);
	if (p%2) return (ans*ans*2ll)%mod;
	else return (ans*ans)%mod;
}

bool line(int p,int q,int r)
{
	return (y[q]-y[p])*(x[r]-x[q])==(y[r]-y[q])*(x[q]-x[p]);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
		scanf("%d%d",&x[i],&y[i]);
	ans=(pw(n)-1ll-(LL)n-(LL)n*(n-1)/2)%mod;
	
	for(int i=1;i<n;i++)
	{
		for(int j=i+1;j<n;j++)
		{
			f=true;
			for(int k=1;k<j;k++)
				if (k!=i && line(i,j,k))
				{
					f=false;
					break;
				}
			if (!f) continue;
			
			int cnt=2;
			for(int k=j+1;k<=n;k++)
				if (line(i,j,k)) cnt++;
			LL tmp=(pw(cnt)-1ll-(LL)cnt-(LL)cnt*(cnt-1)/2)%mod;
			ans-=tmp;
		}
	}
	
	ans=(ans%mod+mod)%mod;
	printf("%I64d\n",ans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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>
using namespace std;
const int maxn=201;
typedef long long LL;
const LL mod=998244353;

int n,x[maxn],y[maxn];
LL ans;
bool f;

LL pw(int p)
{
	if (p==1) return 2ll;
	LL ans=pw(p/2);
	if (p%2) return (ans*ans*2ll)%mod;
	else return (ans*ans)%mod;
}

bool line(int p,int q,int r)
{
	return (y[q]-y[p])*(x[r]-x[q])==(y[r]-y[q])*(x[q]-x[p]);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
		scanf("%d%d",&x[i],&y[i]);
	ans=(pw(n)-1ll-(LL)n-(LL)n*(n-1)/2)%mod;
	
	for(int i=1;i<n;i++)
	{
		for(int j=i+1;j<n;j++)
		{
			f=true;
			for(int k=1;k<j;k++)
				if (k!=i && line(i,j,k))
				{
					f=false;
					break;
				}
			if (!f) continue;
			
			int cnt=2;
			for(int k=j+1;k<=n;k++)
				if (line(i,j,k)) cnt++;
			LL tmp=(pw(cnt)-1ll-(LL)cnt-(LL)cnt*(cnt-1)/2)%mod;
			ans-=tmp;
		}
	}
	
	ans=(ans%mod+mod)%mod;
	printf("%I64d\n",ans);
	return 0;
}
```