#include<bits/stdc++.h>
using namespace std;
struct pnt{
	int x,y;
	pnt(){}
	pnt(int _x,int _y){
		x=_x;y=_y;
	}
};
bool operator <(pnt x,pnt y){
	return x.x < y.x || x.x == y.x && x.y<y.y;
}
pnt operator-(pnt x,pnt y){
	return pnt(x.x-y.x,x.y-y.y);
}
int operator*(pnt x,pnt y){
	return x.x*y.y-x.y*y.x;
}
int n;
pnt a[222];
#define mod 998244353
int add(int x,int y){
	return (x+y)%mod;
}
int sub(int x,int y){
	return add(x,mod-y);
}
int p2[222];
int main(){
	scanf("%d",&n);
	p2[0] = 1;
	for(int i=0; i<n; i++){
		scanf("%d%d",&a[i].x,&a[i].y);
		p2[i+1]=add(p2[i],p2[i]);
	}
	int res = p2[n];
	res = sub(res, 1);
	res = sub(res, n);
	res = sub(res, n*(n-1)/2);
	for(int i=0; i<n; i++)for(int j=0; j<n; j++)
		if(a[i] < a[j]){
			int w = 0;
			for(int k=0; k<n; k++)
			if(a[i] < a[k] && a[j] < a[k] && (a[j]-a[i])*(a[k]-a[i]) == 0)
				w++;
			res = sub(res, sub(p2[w],1));
		}
	printf("%d\n",res);
	return 0;
}

### Prompt

In CPP, your task is to solve 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;
struct pnt{
	int x,y;
	pnt(){}
	pnt(int _x,int _y){
		x=_x;y=_y;
	}
};
bool operator <(pnt x,pnt y){
	return x.x < y.x || x.x == y.x && x.y<y.y;
}
pnt operator-(pnt x,pnt y){
	return pnt(x.x-y.x,x.y-y.y);
}
int operator*(pnt x,pnt y){
	return x.x*y.y-x.y*y.x;
}
int n;
pnt a[222];
#define mod 998244353
int add(int x,int y){
	return (x+y)%mod;
}
int sub(int x,int y){
	return add(x,mod-y);
}
int p2[222];
int main(){
	scanf("%d",&n);
	p2[0] = 1;
	for(int i=0; i<n; i++){
		scanf("%d%d",&a[i].x,&a[i].y);
		p2[i+1]=add(p2[i],p2[i]);
	}
	int res = p2[n];
	res = sub(res, 1);
	res = sub(res, n);
	res = sub(res, n*(n-1)/2);
	for(int i=0; i<n; i++)for(int j=0; j<n; j++)
		if(a[i] < a[j]){
			int w = 0;
			for(int k=0; k<n; k++)
			if(a[i] < a[k] && a[j] < a[k] && (a[j]-a[i])*(a[k]-a[i]) == 0)
				w++;
			res = sub(res, sub(p2[w],1));
		}
	printf("%d\n",res);
	return 0;
}

```