#include <bits/stdc++.h>
using namespace std;
#define N 210
#define ll long long
#define mod 998244353
int n,ans;
struct poi
{
	int x,y;
	poi(){}
	poi(int x,int y):x(x),y(y){}
	friend poi operator - (const poi &r1,const poi &r2)
	{return poi(r1.x-r2.x,r1.y-r2.y);}
	friend int operator ^ (const poi &r1,const poi &r2)
	{return r1.x*r2.y-r2.x*r1.y;}
}a[N];
int qpow(int x,int y)
{
	int ret=1;
	while(y)
	{
		if(y&1)ret=(ll)ret*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}
	return ret;
}
int ni(int x){return qpow(x,mod-2);}
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	ans=(qpow(2,n)-1-n+mod)%mod;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		{
			int num=2;
			for(int k=1;k<=n;k++)
				if(k!=i&&k!=j&&((a[k]-a[i])^(a[k]-a[j]))==0)num++;
			int v1=(qpow(2,num)-num-1+mod)%mod,v2=num*(num-1)/2;
			v1=(ll)v1*ni(v2)%mod;
			ans=(ans-v1+mod)%mod;
		}
	printf("%d\n",ans);
	return 0;
}

### Prompt

Your task is to create 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>
using namespace std;
#define N 210
#define ll long long
#define mod 998244353
int n,ans;
struct poi
{
	int x,y;
	poi(){}
	poi(int x,int y):x(x),y(y){}
	friend poi operator - (const poi &r1,const poi &r2)
	{return poi(r1.x-r2.x,r1.y-r2.y);}
	friend int operator ^ (const poi &r1,const poi &r2)
	{return r1.x*r2.y-r2.x*r1.y;}
}a[N];
int qpow(int x,int y)
{
	int ret=1;
	while(y)
	{
		if(y&1)ret=(ll)ret*x%mod;
		x=(ll)x*x%mod;y>>=1;
	}
	return ret;
}
int ni(int x){return qpow(x,mod-2);}
int main()
{
	//freopen("tt.in","r",stdin);
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	ans=(qpow(2,n)-1-n+mod)%mod;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		{
			int num=2;
			for(int k=1;k<=n;k++)
				if(k!=i&&k!=j&&((a[k]-a[i])^(a[k]-a[j]))==0)num++;
			int v1=(qpow(2,num)-num-1+mod)%mod,v2=num*(num-1)/2;
			v1=(ll)v1*ni(v2)%mod;
			ans=(ans-v1+mod)%mod;
		}
	printf("%d\n",ans);
	return 0;
}

```