#include <bits/stdc++.h>
#define N 210
#define ll long long
#define db double
#define For(i,x,y) for(int i=(x);i<=(y);++i)
#define Rof(i,x,y) for(int i=(x);i>=(y);--i)
#define Edge(x) for(int i=head[x];i;i=e[i].nxt)
#define lson l,mid,o<<1
#define rson mid+1,r,o<<1|1
#define mset(x,y) memset(x,y,sizeof(y))
#define mcpy(x,y) memcpy(x,y,sizeof(x))
#define p_b push_back
#define mod 998244353
using namespace std;
ll _2[N];
int x[N],y[N];
int main(){
	int n;
	scanf("%d",&n);_2[0]=1;
	For(i,1,n) _2[i]=(_2[i-1]+_2[i-1])%mod;
	ll ans=_2[n]-1-n-n*(n-1)/2;
	while(ans<0) ans+=mod;
	For(i,1,n) scanf("%d%d",&x[i],&y[i]);
	For(i,1,n) For(j,i+1,n){
		int tmp=0;
		For(k,j+1,n) if(1ll*(1ll*y[j]-y[i])*(1ll*x[k]-x[j])==1ll*(1ll*y[k]-y[j])*(1ll*x[j]-x[i])) tmp++;
		(ans+=mod+1-_2[tmp])%=mod;
	}
	printf("%lld\n",ans);
}

### 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>
#define N 210
#define ll long long
#define db double
#define For(i,x,y) for(int i=(x);i<=(y);++i)
#define Rof(i,x,y) for(int i=(x);i>=(y);--i)
#define Edge(x) for(int i=head[x];i;i=e[i].nxt)
#define lson l,mid,o<<1
#define rson mid+1,r,o<<1|1
#define mset(x,y) memset(x,y,sizeof(y))
#define mcpy(x,y) memcpy(x,y,sizeof(x))
#define p_b push_back
#define mod 998244353
using namespace std;
ll _2[N];
int x[N],y[N];
int main(){
	int n;
	scanf("%d",&n);_2[0]=1;
	For(i,1,n) _2[i]=(_2[i-1]+_2[i-1])%mod;
	ll ans=_2[n]-1-n-n*(n-1)/2;
	while(ans<0) ans+=mod;
	For(i,1,n) scanf("%d%d",&x[i],&y[i]);
	For(i,1,n) For(j,i+1,n){
		int tmp=0;
		For(k,j+1,n) if(1ll*(1ll*y[j]-y[i])*(1ll*x[k]-x[j])==1ll*(1ll*y[k]-y[j])*(1ll*x[j]-x[i])) tmp++;
		(ans+=mod+1-_2[tmp])%=mod;
	}
	printf("%lld\n",ans);
}

```