#include<cstdio>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn=200+10,mo=998244353;
const db eps=1e-6;
int two[maxn],x[maxn],y[maxn];
int i,j,k,l,t,n,m,ans;
bool check(int i,int j,int k){
	if ((y[j]-y[i])*(x[k]-x[i])==(y[k]-y[i])*(x[j]-x[i])) return 1;else return 0;
}
int main(){
	scanf("%d",&n);
	two[0]=1;
	fo(i,1,n) two[i]=(ll)two[i-1]*2%mo;
	fo(i,1,n) scanf("%d%d",&x[i],&y[i]);
	ans=(two[n]-n-1)%mo;
	fo(i,1,n-1)
		fo(j,i+1,n){
			t=0;
			fo(k,j+1,n)
				if (check(i,j,k)) t++;
			(ans-=two[t])%=mo;
		}
	(ans+=mo)%=mo;
	printf("%d\n",ans);
}
### Prompt

Your challenge is to write 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<cstdio>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn=200+10,mo=998244353;
const db eps=1e-6;
int two[maxn],x[maxn],y[maxn];
int i,j,k,l,t,n,m,ans;
bool check(int i,int j,int k){
	if ((y[j]-y[i])*(x[k]-x[i])==(y[k]-y[i])*(x[j]-x[i])) return 1;else return 0;
}
int main(){
	scanf("%d",&n);
	two[0]=1;
	fo(i,1,n) two[i]=(ll)two[i-1]*2%mo;
	fo(i,1,n) scanf("%d%d",&x[i],&y[i]);
	ans=(two[n]-n-1)%mo;
	fo(i,1,n-1)
		fo(j,i+1,n){
			t=0;
			fo(k,j+1,n)
				if (check(i,j,k)) t++;
			(ans-=two[t])%=mo;
		}
	(ans+=mo)%=mo;
	printf("%d\n",ans);
}
```