#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=210,mod=998244353;
int inc(int x,int y){x+=y;return x>=mod?x-mod:x;}
int dec(int x,int y){x-=y;return x<0?x+mod:x;}
int mul(int x,int y){return (ll)x*y%mod;}
int power(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=mul(x,x))
		if (y&1) ans=mul(ans,x);
	return ans;
}
int calc(int n){
	return dec(power(2,n),n*(n+1)/2+1);
}
typedef pair<int,int> pii;
pii operator - (pii a,pii b){
	return pii(a.first-b.first,a.second-b.second);
}
int operator * (pii a,pii b){
	return a.first*b.second-a.second*b.first;
}
int n,q[N],size;pii p[N];
bool use[N][N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d%d",&p[i].first,&p[i].second);
	int ans=calc(n);
	for (int i=1;i<=n;i++)
	for (int j=1;j<i;j++)
	if (!use[i][j]){
		pii vec=p[i]-p[j];size=0;
		for (int k=1;k<=n;k++)
			if ((p[k]-p[j])*vec==0) q[++size]=k;
		ans=dec(ans,calc(size));
		for (int a=1;a<=size;a++)
		for (int b=1;b<=size;b++)
			use[q[a]][q[b]]=1;
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef long long ll;
const int N=210,mod=998244353;
int inc(int x,int y){x+=y;return x>=mod?x-mod:x;}
int dec(int x,int y){x-=y;return x<0?x+mod:x;}
int mul(int x,int y){return (ll)x*y%mod;}
int power(int x,int y){
	int ans=1;
	for (;y;y>>=1,x=mul(x,x))
		if (y&1) ans=mul(ans,x);
	return ans;
}
int calc(int n){
	return dec(power(2,n),n*(n+1)/2+1);
}
typedef pair<int,int> pii;
pii operator - (pii a,pii b){
	return pii(a.first-b.first,a.second-b.second);
}
int operator * (pii a,pii b){
	return a.first*b.second-a.second*b.first;
}
int n,q[N],size;pii p[N];
bool use[N][N];
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d%d",&p[i].first,&p[i].second);
	int ans=calc(n);
	for (int i=1;i<=n;i++)
	for (int j=1;j<i;j++)
	if (!use[i][j]){
		pii vec=p[i]-p[j];size=0;
		for (int k=1;k<=n;k++)
			if ((p[k]-p[j])*vec==0) q[++size]=k;
		ans=dec(ans,calc(size));
		for (int a=1;a<=size;a++)
		for (int b=1;b<=size;b++)
			use[q[a]][q[b]]=1;
	}
	printf("%d\n",ans);
	return 0;
}
```