#include<bits/stdc++.h>
#define mod 998244353
#define maxn 210
using namespace std;
struct point{
	int x,y;
	point(){}
	point(int x,int y):x(x),y(y){}
	point operator-(const point& p)const{return point(x-p.x,y-p.y);}
}p[maxn];
int n,vis[maxn];
int qpow(int a,int b){
	int ans=1,tmp=a;
	for(;b;b>>=1,tmp=1ll*tmp*tmp%mod)
		if(b&1)ans=1ll*ans*tmp%mod;
	return ans;
}
int cross(point A,point B){
	return A.x*B.y-A.y*B.x;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d%d",&p[i].x,&p[i].y);
	int ans=qpow(2,n);
	for(int i=1;i<=n;++i){
		for(int i=1;i<=n;++i)vis[i]=false;
		for(int j=i+1;j<=n;++j)if(!vis[j]){
			int sum=1;
			for(int k=1;k<i;++k)if(cross(p[k]-p[i],p[j]-p[i])==0)goto nxt;
			for(int k=1;k<=n;++k)if(k!=i&&cross(p[k]-p[i],p[j]-p[i])==0)vis[k]=true,sum++;
			ans=(ans-qpow(2,sum)+sum+1+mod)%mod;
//			printf("[(%d,%d)%d]\n",i,j,sum);
			nxt:;
		}
	}
	ans-=n+1;
	printf("%d",(ans%mod+mod)%mod);
}
### 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<bits/stdc++.h>
#define mod 998244353
#define maxn 210
using namespace std;
struct point{
	int x,y;
	point(){}
	point(int x,int y):x(x),y(y){}
	point operator-(const point& p)const{return point(x-p.x,y-p.y);}
}p[maxn];
int n,vis[maxn];
int qpow(int a,int b){
	int ans=1,tmp=a;
	for(;b;b>>=1,tmp=1ll*tmp*tmp%mod)
		if(b&1)ans=1ll*ans*tmp%mod;
	return ans;
}
int cross(point A,point B){
	return A.x*B.y-A.y*B.x;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
		scanf("%d%d",&p[i].x,&p[i].y);
	int ans=qpow(2,n);
	for(int i=1;i<=n;++i){
		for(int i=1;i<=n;++i)vis[i]=false;
		for(int j=i+1;j<=n;++j)if(!vis[j]){
			int sum=1;
			for(int k=1;k<i;++k)if(cross(p[k]-p[i],p[j]-p[i])==0)goto nxt;
			for(int k=1;k<=n;++k)if(k!=i&&cross(p[k]-p[i],p[j]-p[i])==0)vis[k]=true,sum++;
			ans=(ans-qpow(2,sum)+sum+1+mod)%mod;
//			printf("[(%d,%d)%d]\n",i,j,sum);
			nxt:;
		}
	}
	ans-=n+1;
	printf("%d",(ans%mod+mod)%mod);
}
```