#include<bits/stdc++.h>
#define N 205
using namespace std;
const int mo=998244353;
int n,m,pw[N],ans,cnt;
struct dot{
	int x,y;
	dot(int _x=0,int _y=0){ x=_x,y=_y;}
}a[N];
dot operator +(const dot &p,const dot &q){ return dot(p.x+q.x,p.y+q.y);}
dot operator -(const dot &p,const dot &q){ return dot(p.x-q.x,p.y-q.y);}
int C(int n,int m){ if(m==0) return 1; if(m==1) return n; return n*(n-1)/2;}
int cj(dot p,dot q){ return p.x*q.y-p.y*q.x;}
void inc(int &x,int y){ x=x+y>=mo ? x+y-mo : x+y;}
void dec(int &x,int y){ x=x-y<0 ? x-y+mo : x-y;}
int main()
{
	int i,j,k;
	scanf("%d",&n);
	pw[0]=1;
	for(i=1;i<=n;i++) pw[i]=pw[i-1]*2%mo;
	ans=pw[n];
	dec(ans,C(n,2));
	dec(ans,C(n,1));
	dec(ans,C(n,0));
	for(i=1;i<=n;i++)
		scanf("%d %d",&a[i].x,&a[i].y);
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++){
			for(k=1;k<=j;k++)
				if(k!=i&&k!=j&&cj(a[k]-a[i],a[j]-a[i])==0) break;
			if(k<j) continue;
			cnt=2;
			for(k=j+1;k<=n;k++)
				if(cj(a[k]-a[i],a[j]-a[i])==0) cnt++;
			dec(ans,pw[cnt]);
			inc(ans,C(cnt,0));
			inc(ans,C(cnt,1));
			inc(ans,C(cnt,2));
		  }
	cout<<ans;
	return 0;
}

### Prompt

Please formulate a CPP solution to 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>
#define N 205
using namespace std;
const int mo=998244353;
int n,m,pw[N],ans,cnt;
struct dot{
	int x,y;
	dot(int _x=0,int _y=0){ x=_x,y=_y;}
}a[N];
dot operator +(const dot &p,const dot &q){ return dot(p.x+q.x,p.y+q.y);}
dot operator -(const dot &p,const dot &q){ return dot(p.x-q.x,p.y-q.y);}
int C(int n,int m){ if(m==0) return 1; if(m==1) return n; return n*(n-1)/2;}
int cj(dot p,dot q){ return p.x*q.y-p.y*q.x;}
void inc(int &x,int y){ x=x+y>=mo ? x+y-mo : x+y;}
void dec(int &x,int y){ x=x-y<0 ? x-y+mo : x-y;}
int main()
{
	int i,j,k;
	scanf("%d",&n);
	pw[0]=1;
	for(i=1;i<=n;i++) pw[i]=pw[i-1]*2%mo;
	ans=pw[n];
	dec(ans,C(n,2));
	dec(ans,C(n,1));
	dec(ans,C(n,0));
	for(i=1;i<=n;i++)
		scanf("%d %d",&a[i].x,&a[i].y);
	for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++){
			for(k=1;k<=j;k++)
				if(k!=i&&k!=j&&cj(a[k]-a[i],a[j]-a[i])==0) break;
			if(k<j) continue;
			cnt=2;
			for(k=j+1;k<=n;k++)
				if(cj(a[k]-a[i],a[j]-a[i])==0) cnt++;
			dec(ans,pw[cnt]);
			inc(ans,C(cnt,0));
			inc(ans,C(cnt,1));
			inc(ans,C(cnt,2));
		  }
	cout<<ans;
	return 0;
}

```