#include<bits/stdc++.h>
#define FOR(i,a,b) for (register int i=(a);i<=(b);i++)
#define For(i,a,b) for (register int i=(a);i>=(b);i--)
#define mem(i,j) memset(i,j,sizeof(i))
#define GO(u) for (register int j=f[u];j!=-1;j=nxt[j])
#define fi first
#define se second
#define pii pair<int,int>
#define MP make_pair
#define pb push_back
using namespace std;
typedef long long ll;
const int N=205;
const int mod=998244353;
int n,x[N],y[N],ans=0,er[N];
inline int read()
{
	int x=0,f=1;
	char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
	return f*x;
}
inline void write(int x)
{
	if (x<0) putchar('-'),x=-x;
	if (x>9) write(x/10);
	putchar(x%10+'0');
	return;
}
int main()
{
	n=read();
	FOR(i,1,n) x[i]=read(),y[i]=read();
	er[0]=1;
	FOR(i,1,n) er[i]=2LL*er[i-1]%mod;
	ans=(1LL*er[n]-1-n+mod)%mod;
	FOR(i,1,n) FOR(j,1,i-1) 
	{
		int cnt=0;
		FOR(k,1,j-1) if (1LL*((y[i]-y[j])*(x[j]-x[k]))==1LL*((x[i]-x[j])*(y[j]-y[k]))) cnt++;
		ans=(1LL*ans-er[cnt]+mod)%mod;
	}
	write(ans);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 FOR(i,a,b) for (register int i=(a);i<=(b);i++)
#define For(i,a,b) for (register int i=(a);i>=(b);i--)
#define mem(i,j) memset(i,j,sizeof(i))
#define GO(u) for (register int j=f[u];j!=-1;j=nxt[j])
#define fi first
#define se second
#define pii pair<int,int>
#define MP make_pair
#define pb push_back
using namespace std;
typedef long long ll;
const int N=205;
const int mod=998244353;
int n,x[N],y[N],ans=0,er[N];
inline int read()
{
	int x=0,f=1;
	char c=getchar();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
	while (c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
	return f*x;
}
inline void write(int x)
{
	if (x<0) putchar('-'),x=-x;
	if (x>9) write(x/10);
	putchar(x%10+'0');
	return;
}
int main()
{
	n=read();
	FOR(i,1,n) x[i]=read(),y[i]=read();
	er[0]=1;
	FOR(i,1,n) er[i]=2LL*er[i-1]%mod;
	ans=(1LL*er[n]-1-n+mod)%mod;
	FOR(i,1,n) FOR(j,1,i-1) 
	{
		int cnt=0;
		FOR(k,1,j-1) if (1LL*((y[i]-y[j])*(x[j]-x[k]))==1LL*((x[i]-x[j])*(y[j]-y[k]))) cnt++;
		ans=(1LL*ans-er[cnt]+mod)%mod;
	}
	write(ans);
	return 0;
}
```