#include <complex>
#include <vector>
#include <set>
#include <cstdio>
using namespace std;

const double EPS = 1e-12;
typedef complex<double> P;

long long pow_binary_mod(long long x,long long y,long long M){
	long long z=1;
	for(;y;y>>=1){
		if((y&1)!=0)z=z*x%M;
		x=x*x%M;
	}
	return z;
}

int main(){
	int M=998244353;
	int N,x,y;
	scanf("%d",&N);
	vector<P>v(N);
	int r=pow_binary_mod(2,N,M);
	set<pair<int,int>>se;
	for(int i=0;i<N;i++)scanf("%d%d",&x,&y),v[i]={(double)x,(double)y};
	for(int i=0;i<N;i++)for(int j=i+1;j<N;j++){
		if(se.find({i,j})!=se.end())continue;
		vector<int>x={i,j};
		for(int k=j+1;k<N;k++)if(abs(imag(conj(v[j]-v[k])*(v[i]-v[k])))<1e-12)x.push_back(k);
		for(int a=0;a<x.size();a++)for(int b=a+1;b<x.size();b++){
			r=(r-pow_binary_mod(2,b+~a,M))%M;
			se.emplace(x[a],x[b]);
		}
	}
	printf("%d\n",((r-N-1)%M+M)%M);
}
### 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 <complex>
#include <vector>
#include <set>
#include <cstdio>
using namespace std;

const double EPS = 1e-12;
typedef complex<double> P;

long long pow_binary_mod(long long x,long long y,long long M){
	long long z=1;
	for(;y;y>>=1){
		if((y&1)!=0)z=z*x%M;
		x=x*x%M;
	}
	return z;
}

int main(){
	int M=998244353;
	int N,x,y;
	scanf("%d",&N);
	vector<P>v(N);
	int r=pow_binary_mod(2,N,M);
	set<pair<int,int>>se;
	for(int i=0;i<N;i++)scanf("%d%d",&x,&y),v[i]={(double)x,(double)y};
	for(int i=0;i<N;i++)for(int j=i+1;j<N;j++){
		if(se.find({i,j})!=se.end())continue;
		vector<int>x={i,j};
		for(int k=j+1;k<N;k++)if(abs(imag(conj(v[j]-v[k])*(v[i]-v[k])))<1e-12)x.push_back(k);
		for(int a=0;a<x.size();a++)for(int b=a+1;b<x.size();b++){
			r=(r-pow_binary_mod(2,b+~a,M))%M;
			se.emplace(x[a],x[b]);
		}
	}
	printf("%d\n",((r-N-1)%M+M)%M);
}
```