#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int MOD=998244353;
const int N=200005;

struct pos {int x,y;} p[N];

pos operator -(pos a,pos b) {
	return (pos) {a.x-b.x,a.y-b.y};
}

int operator *(pos a,pos b) {
	return a.x*b.y-b.x*a.y;
}

LL bin[N];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):v,ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

bool line(pos a,pos b,pos c) {
	return (a-b)*(c-b)==0;
}

int main(void) {
	bin[0]=1; rep(i,1,N-1) bin[i]=bin[i-1]*2LL%MOD;
	int n=read();
	rep(i,1,n) p[i].x=read(),p[i].y=read();
	LL ans=bin[n]-n-1-(n-1)*n/2; ans=(ans%MOD+MOD)%MOD;
	rep(i,1,n) rep(j,i+1,n) {
		int s=0;
		rep(k,j+1,n) {
			s+=line(p[i],p[j],p[k]);
		}
		ans=(ans+MOD-bin[s]+1)%MOD;
	}
	printf("%lld\n", ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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 <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int MOD=998244353;
const int N=200005;

struct pos {int x,y;} p[N];

pos operator -(pos a,pos b) {
	return (pos) {a.x-b.x,a.y-b.y};
}

int operator *(pos a,pos b) {
	return a.x*b.y-b.x*a.y;
}

LL bin[N];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):v,ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

bool line(pos a,pos b,pos c) {
	return (a-b)*(c-b)==0;
}

int main(void) {
	bin[0]=1; rep(i,1,N-1) bin[i]=bin[i-1]*2LL%MOD;
	int n=read();
	rep(i,1,n) p[i].x=read(),p[i].y=read();
	LL ans=bin[n]-n-1-(n-1)*n/2; ans=(ans%MOD+MOD)%MOD;
	rep(i,1,n) rep(j,i+1,n) {
		int s=0;
		rep(k,j+1,n) {
			s+=line(p[i],p[j],p[k]);
		}
		ans=(ans+MOD-bin[s]+1)%MOD;
	}
	printf("%lld\n", ans);
	return 0;
}
```