#include<bits/stdc++.h>
const int P=998244353;
struct pos{
	int x,y;
}ps[207];
int n,pw2[207],ans;
struct dir{
	int a,b,z;
	bool operator<(const dir&w)const{return z<w.z;}
}ds[40007];
int dp=0,f[207],sz[207];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int gf(int x){
	while(x!=f[x])x=f[x]=f[f[x]];
	return x;
}
int main(){
	scanf("%d",&n);
	for(int i=pw2[0]=1;i<=n;++i)pw2[i]=pw2[i-1]*2%P;
	for(int i=0;i<=n;++i)pw2[i]=(pw2[i]-i-1)%P;
	ans=pw2[n];
	for(int i=0;i<n;++i)scanf("%d%d",&ps[i].x,&ps[i].y);
	for(int i=0;i<n;++i){
		for(int j=0;j<i;++j){
			int xd=ps[i].x-ps[j].x,yd=ps[i].y-ps[j].y;
			int g=gcd(xd,yd);
			xd/=g,yd/=g;
			if(!xd)yd=1;
			if(!yd)xd=1;
			if(xd<0)xd=-xd,yd=-yd;
			ds[dp++]=(dir){i,j,xd*30000+yd};
		}
	}
	std::sort(ds,ds+dp);
	for(int i=0,j=0;i<dp;i=j){
		for(;j<dp&&ds[i].z==ds[j].z;++j);
		for(int k=0;k<n;++k)f[k]=k,sz[k]=1;
		for(int k=i;k<j;++k){
			int x=gf(ds[k].a),y=gf(ds[k].b);
			if(x!=y)f[x]=y,sz[y]+=sz[x];
		}
		for(int k=0;k<n;++k)if(f[k]==k)ans=(ans-pw2[sz[k]])%P;
	}
	printf("%d\n",(ans%P+P)%P);
	return 0;
}
### Prompt

In CPP, your task is to solve 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>
const int P=998244353;
struct pos{
	int x,y;
}ps[207];
int n,pw2[207],ans;
struct dir{
	int a,b,z;
	bool operator<(const dir&w)const{return z<w.z;}
}ds[40007];
int dp=0,f[207],sz[207];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int gf(int x){
	while(x!=f[x])x=f[x]=f[f[x]];
	return x;
}
int main(){
	scanf("%d",&n);
	for(int i=pw2[0]=1;i<=n;++i)pw2[i]=pw2[i-1]*2%P;
	for(int i=0;i<=n;++i)pw2[i]=(pw2[i]-i-1)%P;
	ans=pw2[n];
	for(int i=0;i<n;++i)scanf("%d%d",&ps[i].x,&ps[i].y);
	for(int i=0;i<n;++i){
		for(int j=0;j<i;++j){
			int xd=ps[i].x-ps[j].x,yd=ps[i].y-ps[j].y;
			int g=gcd(xd,yd);
			xd/=g,yd/=g;
			if(!xd)yd=1;
			if(!yd)xd=1;
			if(xd<0)xd=-xd,yd=-yd;
			ds[dp++]=(dir){i,j,xd*30000+yd};
		}
	}
	std::sort(ds,ds+dp);
	for(int i=0,j=0;i<dp;i=j){
		for(;j<dp&&ds[i].z==ds[j].z;++j);
		for(int k=0;k<n;++k)f[k]=k,sz[k]=1;
		for(int k=i;k<j;++k){
			int x=gf(ds[k].a),y=gf(ds[k].b);
			if(x!=y)f[x]=y,sz[y]+=sz[x];
		}
		for(int k=0;k<n;++k)if(f[k]==k)ans=(ans-pw2[sz[k]])%P;
	}
	printf("%d\n",(ans%P+P)%P);
	return 0;
}
```