#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define mid (l+r>>1)
#define M 202
#define mod 998244353
using namespace std;
LL read(){
	LL nm=0,fh=1;char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
bool vis[M];
LL n,m,C[M][M],X[M],Y[M],T[M],ans;
bool ol(LL t1,LL t2,LL t3){return (Y[t1]-Y[t2])*(X[t1]-X[t3])==(Y[t1]-Y[t3])*(X[t1]-X[t2]);}
int main(){
	memset(vis,false,sizeof(vis));
	n=read(),T[0]=1;
	for(LL i=1;i<=n;i++) T[i]=(T[i-1]<<1)%mod,X[i]=read(),Y[i]=read();
	ans=T[n],ans+=mod-n-1,ans%=mod;
	for(LL i=1;i<=n;i++){
		memset(vis,false,sizeof(vis));
		for(LL j=i+1;j<=n;j++){
			if(vis[j]) continue;
			vis[j]=true;
			LL sum=2;
			bool fg=true;
			for(LL k=1;k<i;k++) if(ol(i,j,k)) fg=false;
			if(!fg) continue;
			for(LL k=j+1;k<=n;k++) if(ol(i,j,k)) sum++,vis[k]=true;
			ans-=T[sum]-sum-1,ans+=mod,ans%=mod;
		}
	}
	printf("%lld\n",ans);
	return 0;
}
### 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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define mid (l+r>>1)
#define M 202
#define mod 998244353
using namespace std;
LL read(){
	LL nm=0,fh=1;char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
bool vis[M];
LL n,m,C[M][M],X[M],Y[M],T[M],ans;
bool ol(LL t1,LL t2,LL t3){return (Y[t1]-Y[t2])*(X[t1]-X[t3])==(Y[t1]-Y[t3])*(X[t1]-X[t2]);}
int main(){
	memset(vis,false,sizeof(vis));
	n=read(),T[0]=1;
	for(LL i=1;i<=n;i++) T[i]=(T[i-1]<<1)%mod,X[i]=read(),Y[i]=read();
	ans=T[n],ans+=mod-n-1,ans%=mod;
	for(LL i=1;i<=n;i++){
		memset(vis,false,sizeof(vis));
		for(LL j=i+1;j<=n;j++){
			if(vis[j]) continue;
			vis[j]=true;
			LL sum=2;
			bool fg=true;
			for(LL k=1;k<i;k++) if(ol(i,j,k)) fg=false;
			if(!fg) continue;
			for(LL k=j+1;k<=n;k++) if(ol(i,j,k)) sum++,vis[k]=true;
			ans-=T[sum]-sum-1,ans+=mod,ans%=mod;
		}
	}
	printf("%lld\n",ans);
	return 0;
}
```