#include <cstdio>
typedef long long ll;

const int N = 203, mod = 998244353;

struct pnt {
	int x, y;
	inline void in() {scanf("%d%d", &x, &y);}
	friend inline pnt operator - (const pnt&A, const pnt&B) {return (pnt) {A.x - B.x, A.y - B.y};}
	friend inline int operator * (const pnt&A, const pnt&B) {return A.x * B.y - A.y * B.x;}
} p[N];

inline bool con(const pnt&A, const pnt&B, const pnt&C) {
	return (C - A) * (B - A) == 0;
}

int n;
int bin[N];
bool vs[N];
ll Ans;

int main() {
	scanf("%d", &n);
	for(int i=bin[0]=1;i<=n;++i) bin[i] = (bin[i-1] << 1)%mod;
	for(int i=1;i<=n;++i) p[i].in();
	for(int i=1;i<n;++i) {
		for(int j=1;j<=n;++j) vs[j] = false;
		
		for(int j=i+1;j<=n;++j)
			if(!vs[j]) {
				int ct = 2;
				for(int k=j+1;k<=n;++k)
					if(con(p[i], p[j], p[k])) vs[k] = true, ++ct;
				
				bool flg = true;
				for(int k=1;k<i and flg;++k)
					if(con(p[i], p[j], p[k])) flg = false;
				if(flg) Ans += bin[ct] - ct - 1;
			}
	}
	Ans = ((ll)mod*mod - Ans - n - 1 + bin[n])%mod;
	printf("%lld\n", 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 <cstdio>
typedef long long ll;

const int N = 203, mod = 998244353;

struct pnt {
	int x, y;
	inline void in() {scanf("%d%d", &x, &y);}
	friend inline pnt operator - (const pnt&A, const pnt&B) {return (pnt) {A.x - B.x, A.y - B.y};}
	friend inline int operator * (const pnt&A, const pnt&B) {return A.x * B.y - A.y * B.x;}
} p[N];

inline bool con(const pnt&A, const pnt&B, const pnt&C) {
	return (C - A) * (B - A) == 0;
}

int n;
int bin[N];
bool vs[N];
ll Ans;

int main() {
	scanf("%d", &n);
	for(int i=bin[0]=1;i<=n;++i) bin[i] = (bin[i-1] << 1)%mod;
	for(int i=1;i<=n;++i) p[i].in();
	for(int i=1;i<n;++i) {
		for(int j=1;j<=n;++j) vs[j] = false;
		
		for(int j=i+1;j<=n;++j)
			if(!vs[j]) {
				int ct = 2;
				for(int k=j+1;k<=n;++k)
					if(con(p[i], p[j], p[k])) vs[k] = true, ++ct;
				
				bool flg = true;
				for(int k=1;k<i and flg;++k)
					if(con(p[i], p[j], p[k])) flg = false;
				if(flg) Ans += bin[ct] - ct - 1;
			}
	}
	Ans = ((ll)mod*mod - Ans - n - 1 + bin[n])%mod;
	printf("%lld\n", Ans);
	return 0;
}
```