#include<bits/stdc++.h>
#define mo 998244353
using namespace std;
int x[200010],y[200010],mi[200010];
set<pair<int,int> >mp;
int main()
{
	int n,w=0;scanf("%d",&n);mi[0]=1;
	for (int i=1;i<=n;i++)
	{
		mi[i]=mi[i-1]*2%mo;
		scanf("%d%d",&x[i],&y[i]);
	}
	w=mi[n];
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++)
	{
		bool err=false;
		for (int k=i+1;k<=n;k++) if (k!=j) if ((x[k]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[k]-y[i])) err=true;
		if (!err)
		{
			int js=0;
			for (int k=1;k<=n;k++) if ((x[k]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[k]-y[i])) js++;
			w=(w-(mi[js]-js-1))%mo;
		}
	}
	printf("%d\n",(w+mo-n-1)%mo);
}
		
### Prompt

Develop a solution in CPP 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<bits/stdc++.h>
#define mo 998244353
using namespace std;
int x[200010],y[200010],mi[200010];
set<pair<int,int> >mp;
int main()
{
	int n,w=0;scanf("%d",&n);mi[0]=1;
	for (int i=1;i<=n;i++)
	{
		mi[i]=mi[i-1]*2%mo;
		scanf("%d%d",&x[i],&y[i]);
	}
	w=mi[n];
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++)
	{
		bool err=false;
		for (int k=i+1;k<=n;k++) if (k!=j) if ((x[k]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[k]-y[i])) err=true;
		if (!err)
		{
			int js=0;
			for (int k=1;k<=n;k++) if ((x[k]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[k]-y[i])) js++;
			w=(w-(mi[js]-js-1))%mo;
		}
	}
	printf("%d\n",(w+mo-n-1)%mo);
}
		
```