#include<cstdio>
#include<algorithm>
using namespace std;
const int N=205,mo=998244353;
int n,x,i,dn,pw[N],ans;
struct arr{int x,y;}a[N],d[N];
int cross(arr A,arr B){return A.x*B.y-B.x*A.y;}
bool operator < (arr A,arr B){return cross(A,B)>0;}
int main(){
	for (scanf("%d",&n),i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	for (pw[0]=1,i=1;i<=n;i++) pw[i]=pw[i-1]*2%mo;
	ans=pw[n]-1-n;
	for (int x=1;x<=n;x++){
		for (dn=0,i=x+1;i<=n;i++) if (i!=x){
			d[++dn]=(arr){a[i].x-a[x].x,a[i].y-a[x].y};
			if (d[dn].x<0) d[dn].x=-d[dn].x,d[dn].y=-d[dn].y;
			if (d[dn].x==0&&d[dn].y>0) d[dn].y=-d[dn].y;
		}
		sort(d+1,d+dn+1);int k=0;
		for (i=1;i<=dn;i++){
			if (!cross(d[i-1],d[i])) k++;
			else ans=(ans-pw[k]+1)%mo,k=1;
		}
		ans=(ans-pw[k]+1)%mo;
	}
	printf("%d",(ans+mo)%mo);
}
### Prompt

Your task is to create 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>
using namespace std;
const int N=205,mo=998244353;
int n,x,i,dn,pw[N],ans;
struct arr{int x,y;}a[N],d[N];
int cross(arr A,arr B){return A.x*B.y-B.x*A.y;}
bool operator < (arr A,arr B){return cross(A,B)>0;}
int main(){
	for (scanf("%d",&n),i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	for (pw[0]=1,i=1;i<=n;i++) pw[i]=pw[i-1]*2%mo;
	ans=pw[n]-1-n;
	for (int x=1;x<=n;x++){
		for (dn=0,i=x+1;i<=n;i++) if (i!=x){
			d[++dn]=(arr){a[i].x-a[x].x,a[i].y-a[x].y};
			if (d[dn].x<0) d[dn].x=-d[dn].x,d[dn].y=-d[dn].y;
			if (d[dn].x==0&&d[dn].y>0) d[dn].y=-d[dn].y;
		}
		sort(d+1,d+dn+1);int k=0;
		for (i=1;i<=dn;i++){
			if (!cross(d[i-1],d[i])) k++;
			else ans=(ans-pw[k]+1)%mo,k=1;
		}
		ans=(ans-pw[k]+1)%mo;
	}
	printf("%d",(ans+mo)%mo);
}
```