#include<cstdio>
#include<algorithm>
#include<cstring>
#define MN 205
using namespace std;
int n;
const int mod=998244353;
struct po{
	int x,y;
	friend po operator-(po x,po y){return (po){x.x-y.x,x.y-y.y};}
	friend int operator*(po x,po y){return x.x*y.x+x.y*y.y;}
	friend int operator/(po x,po y){return x.y*y.x-x.x*y.y;}
}p[MN];
int pow(int x,int y){
	int tmp=1;
	while(y){
		if(y&1)tmp=1ll*tmp*x%mod;y>>=1;x=1ll*x*x%mod;
	}return tmp;
}
void dec(long long &x,int y){(x-=y)<0?x+=mod:0;}
long long ans=0;
int main(){
	scanf("%d",&n);ans=pow(2,n);dec(ans,n+1);
	for(int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i].y);
	for(int i=1;i<=n;i++)
		for(int j=i+1,cnt=0;j<=n;j++,dec(ans,pow(2,cnt)),cnt=0)
			for(int k=1;k<=n;k++)if((p[k]-p[i])/(p[k]-p[j])==0&&(p[k]-p[i])*(p[k]-p[j])<0)cnt++;
	printf("%lld\n",ans);
}


### Prompt

Create a solution in cpp for 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<cstdio>
#include<algorithm>
#include<cstring>
#define MN 205
using namespace std;
int n;
const int mod=998244353;
struct po{
	int x,y;
	friend po operator-(po x,po y){return (po){x.x-y.x,x.y-y.y};}
	friend int operator*(po x,po y){return x.x*y.x+x.y*y.y;}
	friend int operator/(po x,po y){return x.y*y.x-x.x*y.y;}
}p[MN];
int pow(int x,int y){
	int tmp=1;
	while(y){
		if(y&1)tmp=1ll*tmp*x%mod;y>>=1;x=1ll*x*x%mod;
	}return tmp;
}
void dec(long long &x,int y){(x-=y)<0?x+=mod:0;}
long long ans=0;
int main(){
	scanf("%d",&n);ans=pow(2,n);dec(ans,n+1);
	for(int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i].y);
	for(int i=1;i<=n;i++)
		for(int j=i+1,cnt=0;j<=n;j++,dec(ans,pow(2,cnt)),cnt=0)
			for(int k=1;k<=n;k++)if((p[k]-p[i])/(p[k]-p[j])==0&&(p[k]-p[i])*(p[k]-p[j])<0)cnt++;
	printf("%lld\n",ans);
}


```