#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

#define pb push_back
#define fr first
#define sc second

const ll M = 998244353;

ll modpow(ll x,ll k){
	if(k == 0)return 1;
	ll ret = modpow(x,k/2);
	ret *= ret; ret %= M;
	if(k%2 == 1){
		ret *= x;
		ret %= M;
	}
	return ret;
}

int main(){
	int N;
	int x[202],y[202];
	scanf("%d",&N);
	for(int i = 1 ; i <= N ; i ++){
		scanf("%d%d",&x[i],&y[i]);
	}
	
	ll ret = modpow(2,N);
	ret += M-N-1; if(ret>=M)ret -= M;
	
	for(int i = 1 ; i <= N ; i ++){
		for(int j = i+1 ; j <= N ; j ++){
			ll cnt = 0;
			for(int k = 1 ; k <= N ; k ++){
				if(k == i || k == j)continue;
				if(x[k] < min(x[i],x[j]))continue;
				if(x[k] > max(x[i],x[j]))continue;
				if(y[k] < min(y[i],y[j]))continue;
				if(y[k] > max(y[i],y[j]))continue;
				if((x[k]-x[i])*(y[j]-y[i])-(y[k]-y[i])*(x[j]-x[i]) == 0)cnt ++;
			}
			ret += M-modpow(2,cnt); if(ret>=M)ret -= M;
		}
	}
	cout << ret << endl;
}
	


### Prompt

Construct a cpp code solution to the problem outlined:

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<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

#define pb push_back
#define fr first
#define sc second

const ll M = 998244353;

ll modpow(ll x,ll k){
	if(k == 0)return 1;
	ll ret = modpow(x,k/2);
	ret *= ret; ret %= M;
	if(k%2 == 1){
		ret *= x;
		ret %= M;
	}
	return ret;
}

int main(){
	int N;
	int x[202],y[202];
	scanf("%d",&N);
	for(int i = 1 ; i <= N ; i ++){
		scanf("%d%d",&x[i],&y[i]);
	}
	
	ll ret = modpow(2,N);
	ret += M-N-1; if(ret>=M)ret -= M;
	
	for(int i = 1 ; i <= N ; i ++){
		for(int j = i+1 ; j <= N ; j ++){
			ll cnt = 0;
			for(int k = 1 ; k <= N ; k ++){
				if(k == i || k == j)continue;
				if(x[k] < min(x[i],x[j]))continue;
				if(x[k] > max(x[i],x[j]))continue;
				if(y[k] < min(y[i],y[j]))continue;
				if(y[k] > max(y[i],y[j]))continue;
				if((x[k]-x[i])*(y[j]-y[i])-(y[k]-y[i])*(x[j]-x[i]) == 0)cnt ++;
			}
			ret += M-modpow(2,cnt); if(ret>=M)ret -= M;
		}
	}
	cout << ret << endl;
}
	


```