#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int MX = 205, MOD = 998244353;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, x[MX], y[MX],u;
ll ans=1;
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%d%d", x+i, y+i);
	u = 1+n;
	rep(i,n) for(int j=i+1; j<n; j++){
		ll t=1, tx = x[j]-x[i], ty = y[j]-y[i];
		for(int k=j+1; k<n; k++) if(tx*(y[k]-y[i])-ty*(x[k]-x[i])==0) t = t*2%MOD;
		u = (u+t) % MOD;
	}
	for(ll b=2; n; n/=2, (b*=b)%=MOD) if(n&1) (ans *= b) %= MOD;
	printf("%lld\n", (ans+MOD-u)%MOD);
}
### Prompt

Generate 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;
using ll = long long;
const int MX = 205, MOD = 998244353;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, x[MX], y[MX],u;
ll ans=1;
int main(){
	scanf("%d", &n);
	rep(i,n) scanf("%d%d", x+i, y+i);
	u = 1+n;
	rep(i,n) for(int j=i+1; j<n; j++){
		ll t=1, tx = x[j]-x[i], ty = y[j]-y[i];
		for(int k=j+1; k<n; k++) if(tx*(y[k]-y[i])-ty*(x[k]-x[i])==0) t = t*2%MOD;
		u = (u+t) % MOD;
	}
	for(ll b=2; n; n/=2, (b*=b)%=MOD) if(n&1) (ans *= b) %= MOD;
	printf("%lld\n", (ans+MOD-u)%MOD);
}
```