#include<bits/stdc++.h>
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
typedef pair<pi,pi> ppi;

const int N=205,P=998244353;

map<ppi,int>D;
int n,x[N],y[N],p2[N],m,t;

inline int gcd(int x,int y){return y?gcd(y,x%y):x;}
inline pi mk(pi x){
	ll g=gcd(x.fi,x.se);
	if(g)x.fi/=g,x.se/=g;
	if(x.fi<0)x.fi=-x.fi,x.se=-x.se;
	else if(x.fi==0&&x.se<0)x.se=-x.se;
	return x;
}
inline ppi mp(ll u1,ll v1,ll u2,ll v2){
	pi l=mk({u2-u1,v2-v1}),s;
	if(l.fi)s=mk({v1*l.fi-u1*l.se,l.fi});else s={u1,0};
	return {l,s};
}
inline int get(int s){return (1+sqrt(1+8*s+0.5))/2;}

int main(){
	scanf("%d",&n);
	p2[0]=1;
	for(int i=1;i<=n;i++)p2[i]=(p2[i-1]<<1)%P,scanf("%d%d",&x[i],&y[i]);
	for(int i=1;i<=n;i++)for(int j=1;j<i;j++)D[mp(x[i],y[i],x[j],y[j])]++;
	m=p2[n];
	for(auto&i:D)t=get(i.se),m=(m-p2[t]+t+1+P)%P;
	m=(m-n-1+P)%P;
	printf("%d\n",m);
	return 0;
}
### 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<bits/stdc++.h>
#define fi first
#define se second

using namespace std;
typedef long long ll;
typedef pair<ll,ll> pi;
typedef pair<pi,pi> ppi;

const int N=205,P=998244353;

map<ppi,int>D;
int n,x[N],y[N],p2[N],m,t;

inline int gcd(int x,int y){return y?gcd(y,x%y):x;}
inline pi mk(pi x){
	ll g=gcd(x.fi,x.se);
	if(g)x.fi/=g,x.se/=g;
	if(x.fi<0)x.fi=-x.fi,x.se=-x.se;
	else if(x.fi==0&&x.se<0)x.se=-x.se;
	return x;
}
inline ppi mp(ll u1,ll v1,ll u2,ll v2){
	pi l=mk({u2-u1,v2-v1}),s;
	if(l.fi)s=mk({v1*l.fi-u1*l.se,l.fi});else s={u1,0};
	return {l,s};
}
inline int get(int s){return (1+sqrt(1+8*s+0.5))/2;}

int main(){
	scanf("%d",&n);
	p2[0]=1;
	for(int i=1;i<=n;i++)p2[i]=(p2[i-1]<<1)%P,scanf("%d%d",&x[i],&y[i]);
	for(int i=1;i<=n;i++)for(int j=1;j<i;j++)D[mp(x[i],y[i],x[j],y[j])]++;
	m=p2[n];
	for(auto&i:D)t=get(i.se),m=(m-p2[t]+t+1+P)%P;
	m=(m-n-1+P)%P;
	printf("%d\n",m);
	return 0;
}
```