#include <bits/stdc++.h>
#define ll long long
#define lf double
#define E complex<lf>
#define inf 0x3f3f3f3f
#define eps 1e-8
#define pa pair<int,int>
#define pb push_back
#define ms(x,y) memset(x,y,sizeof(x))
#define l(x) (x<<1)
#define r(x) (x<<1|1)
#define mod 998244353
#define N 210
using namespace std;
inline ll read() {
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
int n,bin[N];
struct hhh {
	int x,y;
}p[N];
int main() {
	n=read(),bin[0]=1;
	for (int i=1; i<=n; i++) p[i].x=read(),p[i].y=read(),bin[i]=bin[i-1]*2%mod;
	int ans=bin[n]-n-1;
	for (int i=1; i<n; i++) 
		for (int j=i+1; j<=n; j++) {
			int cnt=0;
			for (int k=j+1; k<=n; k++) if ((p[i].y-p[k].y)*(p[j].x-p[k].x)==(p[j].y-p[k].y)*(p[i].x-p[k].x)) cnt++;
//			cout << cnt << endl;
			(ans-=bin[cnt])%=mod;
		}
	ans=(ans+mod)%mod;
	cout << ans << endl;
}
### Prompt

Please formulate 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 ll long long
#define lf double
#define E complex<lf>
#define inf 0x3f3f3f3f
#define eps 1e-8
#define pa pair<int,int>
#define pb push_back
#define ms(x,y) memset(x,y,sizeof(x))
#define l(x) (x<<1)
#define r(x) (x<<1|1)
#define mod 998244353
#define N 210
using namespace std;
inline ll read() {
	ll x=0,f=1;char c=getchar();
	while (c<'0'||c>'9') f=(c=='-')?-1:1,c=getchar();
	while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
	return x*f;
}
int n,bin[N];
struct hhh {
	int x,y;
}p[N];
int main() {
	n=read(),bin[0]=1;
	for (int i=1; i<=n; i++) p[i].x=read(),p[i].y=read(),bin[i]=bin[i-1]*2%mod;
	int ans=bin[n]-n-1;
	for (int i=1; i<n; i++) 
		for (int j=i+1; j<=n; j++) {
			int cnt=0;
			for (int k=j+1; k<=n; k++) if ((p[i].y-p[k].y)*(p[j].x-p[k].x)==(p[j].y-p[k].y)*(p[i].x-p[k].x)) cnt++;
//			cout << cnt << endl;
			(ans-=bin[cnt])%=mod;
		}
	ans=(ans+mod)%mod;
	cout << ans << endl;
}
```