#include<cstdio>
#include<algorithm>
using namespace std;
const int N=220,P=998244353;
int i,j,k,n,m,ch,ans,g;
int tm2[N];
struct point {
	int x,y;
	bool operator < (const point &n) const {
		if (x==n.x) return y<n.y;
		return x<n.x;
	}
} A[N],B[N];
void R(int &x) {
	x=0;ch=getchar();
	while (ch<'0' || '9'<ch) ch=getchar();
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
}
int gcd(int a,int b) {
	if (!b) return a;
	return gcd(b,a%b);
}
int main() {
	R(n);
	tm2[0]=1;
	for (i=1;i<=n;i++) tm2[i]=(tm2[i-1]+tm2[i-1])%P;
	for (i=1;i<=n;i++) R(A[i].x),R(A[i].y);
	sort(A+1,A+n+1);
	ans=1;
	for (i=1;i<=n;i++) {
		m=0;
		ans=(ans+1)%P;
		for (j=i+1;j<=n;j++) {
			m++;
			B[m].x=A[j].x-A[i].x;
			B[m].y=A[j].y-A[i].y;
			if (B[m].y<0) g=gcd(B[m].x,-B[m].y);
			else g=gcd(B[m].x,B[m].y);
			B[m].x/=g;
			B[m].y/=g;
		}
		sort(B+1,B+m+1);
		B[m+1].x=-1;
		k=1;
		for (j=1;j<=m;j++) {
			if (B[j].x==B[j+1].x && B[j].y==B[j+1].y) k++;
			else {
				ans=(ans+tm2[k]-1)%P;
				k=1;
			}
		}
	}
	ans=(tm2[n]+P-ans)%P;
	printf("%d\n",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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=220,P=998244353;
int i,j,k,n,m,ch,ans,g;
int tm2[N];
struct point {
	int x,y;
	bool operator < (const point &n) const {
		if (x==n.x) return y<n.y;
		return x<n.x;
	}
} A[N],B[N];
void R(int &x) {
	x=0;ch=getchar();
	while (ch<'0' || '9'<ch) ch=getchar();
	while ('0'<=ch && ch<='9') x=x*10+ch-'0',ch=getchar();
}
int gcd(int a,int b) {
	if (!b) return a;
	return gcd(b,a%b);
}
int main() {
	R(n);
	tm2[0]=1;
	for (i=1;i<=n;i++) tm2[i]=(tm2[i-1]+tm2[i-1])%P;
	for (i=1;i<=n;i++) R(A[i].x),R(A[i].y);
	sort(A+1,A+n+1);
	ans=1;
	for (i=1;i<=n;i++) {
		m=0;
		ans=(ans+1)%P;
		for (j=i+1;j<=n;j++) {
			m++;
			B[m].x=A[j].x-A[i].x;
			B[m].y=A[j].y-A[i].y;
			if (B[m].y<0) g=gcd(B[m].x,-B[m].y);
			else g=gcd(B[m].x,B[m].y);
			B[m].x/=g;
			B[m].y/=g;
		}
		sort(B+1,B+m+1);
		B[m+1].x=-1;
		k=1;
		for (j=1;j<=m;j++) {
			if (B[j].x==B[j+1].x && B[j].y==B[j+1].y) k++;
			else {
				ans=(ans+tm2[k]-1)%P;
				k=1;
			}
		}
	}
	ans=(tm2[n]+P-ans)%P;
	printf("%d\n",ans);
}
```