#include <cstdio>
#include <cstring>
#define MAXN 210
#define LL long long

const LL P=998244353;

LL getPow(LL x,LL y){
	LL res=1;
	while(y){
		if(y&1) res=res*x%P;
		x=x*x%P;
		y>>=1;
	}
	return res;
}

struct vec2{
	int x,y;
	vec2(int _x=0,int _y=0):x(_x),y(_y){}
	friend vec2 operator-(vec2 x,vec2 y){ return vec2(x.x-y.x,x.y-y.y); }
	friend int dot(vec2 x,vec2 y){ return x.x*y.x+x.y*y.y; }
	friend int cross(vec2 x,vec2 y){ return x.x*y.y-x.y*y.x; }
}p[MAXN];

int n;

int main(){
#ifdef DEBUG
	freopen("E.in","r",stdin);
#endif
	LL inv2=getPow(2,P-2);
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d%d",&p[i].x,&p[i].y);
	LL ans=getPow(2,n);
	LL res=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(i^j){
				vec2 d=p[j]-p[i];
				int cnt=2;
				for(int k=1;k<=n;k++)
					if(cross(p[k]-p[j],d)==0 && k!=i && k!=j){
						if(dot(p[k]-p[j],d)<0){
							cnt=0;
							break;
						}
						cnt++;
					}
				LL temp=getPow(2,cnt)-cnt-1;
				res=(res+temp)%P;
			}
	res=res*inv2%P;
	res=(res+n+1)%P;
	ans-=res;
	ans=(ans%P+P)%P;
	printf("%lld\n",ans);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <cstring>
#define MAXN 210
#define LL long long

const LL P=998244353;

LL getPow(LL x,LL y){
	LL res=1;
	while(y){
		if(y&1) res=res*x%P;
		x=x*x%P;
		y>>=1;
	}
	return res;
}

struct vec2{
	int x,y;
	vec2(int _x=0,int _y=0):x(_x),y(_y){}
	friend vec2 operator-(vec2 x,vec2 y){ return vec2(x.x-y.x,x.y-y.y); }
	friend int dot(vec2 x,vec2 y){ return x.x*y.x+x.y*y.y; }
	friend int cross(vec2 x,vec2 y){ return x.x*y.y-x.y*y.x; }
}p[MAXN];

int n;

int main(){
#ifdef DEBUG
	freopen("E.in","r",stdin);
#endif
	LL inv2=getPow(2,P-2);
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d%d",&p[i].x,&p[i].y);
	LL ans=getPow(2,n);
	LL res=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			if(i^j){
				vec2 d=p[j]-p[i];
				int cnt=2;
				for(int k=1;k<=n;k++)
					if(cross(p[k]-p[j],d)==0 && k!=i && k!=j){
						if(dot(p[k]-p[j],d)<0){
							cnt=0;
							break;
						}
						cnt++;
					}
				LL temp=getPow(2,cnt)-cnt-1;
				res=(res+temp)%P;
			}
	res=res*inv2%P;
	res=(res+n+1)%P;
	ans-=res;
	ans=(ans%P+P)%P;
	printf("%lld\n",ans);
}

```