#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
const int mod=998244353;
int n,st[210],top;
bool vis[210][210];
ll ans,cpow[210];
struct pt
{
	int x,y;
}a[210];
bool check(int p,int q,int r)
{
	if((a[p].x-a[q].x)*(a[q].y-a[r].y)==(a[q].x-a[r].x)*(a[p].y-a[q].y)) return 1;
	return 0;
}
int main()
{
	scanf("%d",&n);
	cpow[0]=1;
	for(int i=1;i<=n;i++) cpow[i]=(cpow[i-1]<<1)%mod;
	
	ans=cpow[n];
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
		for(int j=1;j<=i-1;j++)
			if(!vis[i][j])
			{
				top=2;st[1]=i;st[2]=j;
				for(int k=1;k<=n;k++)
					if(k!=i&&k!=j&&check(i,j,k)) st[++top]=k;
				ans=(ans-cpow[top]+top+1+mod)%mod;	
					
				for(int k=1;k<=top;k++)
					for(int l=1;l<=k-1;l++)
						vis[st[k]][st[l]]=vis[st[l]][st[k]]=1;
			}
	ans=(ans-n-1+mod)%mod;
	printf("%lld",ans);	
	return 0;
}
/*5
0 4
1 2
0 0
2 4
2 0*/
### Prompt

Please create a solution in CPP 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>
#include<cstring>
#include<algorithm>
#include<cmath>
#define ll long long
using namespace std;
const int mod=998244353;
int n,st[210],top;
bool vis[210][210];
ll ans,cpow[210];
struct pt
{
	int x,y;
}a[210];
bool check(int p,int q,int r)
{
	if((a[p].x-a[q].x)*(a[q].y-a[r].y)==(a[q].x-a[r].x)*(a[p].y-a[q].y)) return 1;
	return 0;
}
int main()
{
	scanf("%d",&n);
	cpow[0]=1;
	for(int i=1;i<=n;i++) cpow[i]=(cpow[i-1]<<1)%mod;
	
	ans=cpow[n];
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
		for(int j=1;j<=i-1;j++)
			if(!vis[i][j])
			{
				top=2;st[1]=i;st[2]=j;
				for(int k=1;k<=n;k++)
					if(k!=i&&k!=j&&check(i,j,k)) st[++top]=k;
				ans=(ans-cpow[top]+top+1+mod)%mod;	
					
				for(int k=1;k<=top;k++)
					for(int l=1;l<=k-1;l++)
						vis[st[k]][st[l]]=vis[st[l]][st[k]]=1;
			}
	ans=(ans-n-1+mod)%mod;
	printf("%lld",ans);	
	return 0;
}
/*5
0 4
1 2
0 0
2 4
2 0*/
```