#include<bits/stdc++.h>
#define N 205
using namespace std;

#define db double
#define eps 1e-8
struct point{db x,y;};
point operator +(point a,point b){return (point){a.x+b.x,a.y+b.y};}
point operator -(point a,point b){return (point){a.x-b.x,a.y-b.y};}
db operator *(point a,point b){return a.x*b.y-b.x*a.y;}
point operator *(point a,db x){return (point){a.x*x,a.y*x};}
point operator /(point a,db x){return (point){a.x/x,a.y/x};}
bool a_onleft(point a,point b){return a*b<-eps;}
db area(point a,point b,point c){return abs((a-c)*(b-c)/2.0);}
point inter(point a1,point a2,point b1,point b2){
	double s1=area(b1,b2,a1),s2=area(b1,b2,a2);
	return (a1-a2)/(s1+s2)*s2+a2;
}
#undef db
#undef eps
const int mod=998244353;
point p[N];
int n,pw[N];
int main(){
	scanf("%d",&n);pw[0]=1;
	for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y),pw[i]=pw[i-1]*2%mod;
	long long ans=pw[n]-1-n;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++){
			int cnt=0;
			for(int k=j+1;k<=n;k++)
				if(area(p[i],p[j],p[k])<1e-8) cnt++;
			ans-=pw[cnt];
		}
	cout<<(ans%mod+mod)%mod<<endl;
	return 0;
}

### Prompt

Generate 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<bits/stdc++.h>
#define N 205
using namespace std;

#define db double
#define eps 1e-8
struct point{db x,y;};
point operator +(point a,point b){return (point){a.x+b.x,a.y+b.y};}
point operator -(point a,point b){return (point){a.x-b.x,a.y-b.y};}
db operator *(point a,point b){return a.x*b.y-b.x*a.y;}
point operator *(point a,db x){return (point){a.x*x,a.y*x};}
point operator /(point a,db x){return (point){a.x/x,a.y/x};}
bool a_onleft(point a,point b){return a*b<-eps;}
db area(point a,point b,point c){return abs((a-c)*(b-c)/2.0);}
point inter(point a1,point a2,point b1,point b2){
	double s1=area(b1,b2,a1),s2=area(b1,b2,a2);
	return (a1-a2)/(s1+s2)*s2+a2;
}
#undef db
#undef eps
const int mod=998244353;
point p[N];
int n,pw[N];
int main(){
	scanf("%d",&n);pw[0]=1;
	for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y),pw[i]=pw[i-1]*2%mod;
	long long ans=pw[n]-1-n;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++){
			int cnt=0;
			for(int k=j+1;k<=n;k++)
				if(area(p[i],p[j],p[k])<1e-8) cnt++;
			ans-=pw[cnt];
		}
	cout<<(ans%mod+mod)%mod<<endl;
	return 0;
}

```