#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){
	int x=0,f=0;
	char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) f^=(ch=='-');
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return f ? -x : x;
}
const int P=998244353,maxn=1010;
inline int kasumi(int x,int y) {
	int qwq=1;
	while (y) {
		if (y&1) qwq=qwq*x%P;
		y>>=1;
		x=x*x%P;
	}
	return qwq;
}
int x[maxn],y[maxn],n,ans;
signed main(){
	n=read();
	for (int i=1;i<=n;++i) {
		x[i]=read();y[i]=read();
	} 
	ans=(kasumi(2,n)-n-1-n*(n-1)/2+P)%P;
	for (int i=1;i<=n;++i) {
		for (int j=i+1;j<=n;++j) {
			int qwq=0;
			for (int k=j+1;k<=n;++k) {
				if ((x[i]-x[j])*(y[i]-y[k])==(x[i]-x[k])*(y[i]-y[j])) ++qwq;
			}
			ans=(ans-kasumi(2,qwq)+P+1)%P;
		}
	}
	printf("%lld\n",ans);
	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>
using namespace std;
#define int long long
inline int read(){
	int x=0,f=0;
	char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar()) f^=(ch=='-');
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return f ? -x : x;
}
const int P=998244353,maxn=1010;
inline int kasumi(int x,int y) {
	int qwq=1;
	while (y) {
		if (y&1) qwq=qwq*x%P;
		y>>=1;
		x=x*x%P;
	}
	return qwq;
}
int x[maxn],y[maxn],n,ans;
signed main(){
	n=read();
	for (int i=1;i<=n;++i) {
		x[i]=read();y[i]=read();
	} 
	ans=(kasumi(2,n)-n-1-n*(n-1)/2+P)%P;
	for (int i=1;i<=n;++i) {
		for (int j=i+1;j<=n;++j) {
			int qwq=0;
			for (int k=j+1;k<=n;++k) {
				if ((x[i]-x[j])*(y[i]-y[k])==(x[i]-x[k])*(y[i]-y[j])) ++qwq;
			}
			ans=(ans-kasumi(2,qwq)+P+1)%P;
		}
	}
	printf("%lld\n",ans);
	return 0;
}
```