#include<iostream>
#include<cstdio>
using namespace std;
int mod=998244353;
int x[101000],y[101000];
int ans=0;
long long Pow(long long x,int p)
{
	long long ans=1;
	while(p)
	{
		if(p&1)ans=(ans*x)%mod;
		x=(x*x)%mod;p>>=1;
	}
	return ans;
}
int getans(int x){return (Pow(2,x) - ( (1 + x + x * ( x - 1 ) / 2) % mod) + mod ) % mod;}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
	int ans=getans(n);
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			int cnt=0;
			for(int k=1;k<=n;k++)
			{
				if(k==i||k==j)continue;
				if((x[k] - x[i]) * (y[k] - y[j]) != (x[k] - x[j]) * (y[k] - y[i]))continue;
				if(k<j)break;
				cnt++;
			}
			ans=(ans-getans(cnt+2)+mod)%mod;
		}
	}
	printf("%d",ans);return 0;
 } 
### Prompt

Your challenge is to write 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<iostream>
#include<cstdio>
using namespace std;
int mod=998244353;
int x[101000],y[101000];
int ans=0;
long long Pow(long long x,int p)
{
	long long ans=1;
	while(p)
	{
		if(p&1)ans=(ans*x)%mod;
		x=(x*x)%mod;p>>=1;
	}
	return ans;
}
int getans(int x){return (Pow(2,x) - ( (1 + x + x * ( x - 1 ) / 2) % mod) + mod ) % mod;}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
	int ans=getans(n);
	for(int i=1;i<=n;i++)
	{
		for(int j=i+1;j<=n;j++)
		{
			int cnt=0;
			for(int k=1;k<=n;k++)
			{
				if(k==i||k==j)continue;
				if((x[k] - x[i]) * (y[k] - y[j]) != (x[k] - x[j]) * (y[k] - y[i]))continue;
				if(k<j)break;
				cnt++;
			}
			ans=(ans-getans(cnt+2)+mod)%mod;
		}
	}
	printf("%d",ans);return 0;
 } 
```