#include<bits/stdc++.h>
using namespace std;
const int N = 200;
const int mod = 998244353;
int x[N],y[N];
typedef pair<int,int> pii;
int C(int i,int j){
	return x[i]*y[j]-x[j]*y[i];
}
int ccw(int i,int j,int k){
	return C(i,j)+C(j,k)+C(k,i);
}
int pp[N+1];
#define mul(x,y) ((long long)(x)*(y)%mod)
int main(){
	int n;scanf("%d",&n);
	pp[0]=1;for(int i=1;i<=n;i++)pp[i]=mul(pp[i-1],2);
	for(int i=0;i<n;i++){
		scanf("%d%d",x+i,y+i);
	}
	int ans=pp[n];
	ans = (ans+mod-1)%mod;
	ans = (ans+mod-n)%mod;
	for(int i=0;i<n;i++){
		for(int j=0;j<i;j++){
			int r=n;
			pii a = min(pii(x[i],y[i]),pii(x[j],y[j]));
			pii b = max(pii(x[i],y[i]),pii(x[j],y[j]));
			for(int k=0;k<n;k++){
				if(ccw(i,j,k)==0 && a <= pii(x[k],y[k]) && pii(x[k],y[k])<=b)r--;
			}
			ans-=pp[n-2-r];
			if(ans<0)ans+=mod;
		}
	}
	printf("%d\n",ans);
}
### Prompt

Please create a solution in CPP 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;
const int N = 200;
const int mod = 998244353;
int x[N],y[N];
typedef pair<int,int> pii;
int C(int i,int j){
	return x[i]*y[j]-x[j]*y[i];
}
int ccw(int i,int j,int k){
	return C(i,j)+C(j,k)+C(k,i);
}
int pp[N+1];
#define mul(x,y) ((long long)(x)*(y)%mod)
int main(){
	int n;scanf("%d",&n);
	pp[0]=1;for(int i=1;i<=n;i++)pp[i]=mul(pp[i-1],2);
	for(int i=0;i<n;i++){
		scanf("%d%d",x+i,y+i);
	}
	int ans=pp[n];
	ans = (ans+mod-1)%mod;
	ans = (ans+mod-n)%mod;
	for(int i=0;i<n;i++){
		for(int j=0;j<i;j++){
			int r=n;
			pii a = min(pii(x[i],y[i]),pii(x[j],y[j]));
			pii b = max(pii(x[i],y[i]),pii(x[j],y[j]));
			for(int k=0;k<n;k++){
				if(ccw(i,j,k)==0 && a <= pii(x[k],y[k]) && pii(x[k],y[k])<=b)r--;
			}
			ans-=pp[n-2-r];
			if(ans<0)ans+=mod;
		}
	}
	printf("%d\n",ans);
}
```