#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; 
	cin >> N; 
	long long int X[210] = {}; 
	long long int Y[210] = {}; 
	for(int i = 0; i < N; i++){
		cin >> X[i] >> Y[i]; 
	}
	long long int P = 998244353; 
	long long int A[210] = {}; // 一般項2^n-n-1
	bool B[210][210] = {}; 
	for(int i = 1; i <= 200; i++){
		A[i] = (2 * A[i-1] + i - 1) % P; 
	}
	long long int ans = A[N]; 
	for(int i = 0; i < N; i++){ //2点以下の集合も含めて和をとると，各部分集合が1回ずつ数えられる(ように重みが設定されている)ので総和は2^N．2点以下の分を引く．0点は1，1点はN，2点はちょうどm点が一直線上とすると2^m-m-1で，これをすべての直線について加える．
		for(int j = i+1; j < N; j++){
			if(B[i][j] == true) continue;
			vector<int> tansaku; 
			tansaku.push_back(i); 
			tansaku.push_back(j); 
			for(int k = j+1; k < N; k++){
				long long int s = (Y[i]-Y[j])*X[k] + (X[j]-X[i])*Y[k] - (Y[i]*X[j]-X[i]*Y[j]);
				if(s == 0){
					tansaku.push_back(k); 
				}
			}
		int c = tansaku.size(); 
		for(int p = 0; p < c; p++){
			for(int q = p+1; q < c; q++){
				B[tansaku.at(p)][tansaku.at(q)] = true; 
			}
		}
          	ans = (ans + P - A[c]) % P; 
		}
	}
	cout << ans << endl; 
}
### 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 <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
	int N; 
	cin >> N; 
	long long int X[210] = {}; 
	long long int Y[210] = {}; 
	for(int i = 0; i < N; i++){
		cin >> X[i] >> Y[i]; 
	}
	long long int P = 998244353; 
	long long int A[210] = {}; // 一般項2^n-n-1
	bool B[210][210] = {}; 
	for(int i = 1; i <= 200; i++){
		A[i] = (2 * A[i-1] + i - 1) % P; 
	}
	long long int ans = A[N]; 
	for(int i = 0; i < N; i++){ //2点以下の集合も含めて和をとると，各部分集合が1回ずつ数えられる(ように重みが設定されている)ので総和は2^N．2点以下の分を引く．0点は1，1点はN，2点はちょうどm点が一直線上とすると2^m-m-1で，これをすべての直線について加える．
		for(int j = i+1; j < N; j++){
			if(B[i][j] == true) continue;
			vector<int> tansaku; 
			tansaku.push_back(i); 
			tansaku.push_back(j); 
			for(int k = j+1; k < N; k++){
				long long int s = (Y[i]-Y[j])*X[k] + (X[j]-X[i])*Y[k] - (Y[i]*X[j]-X[i]*Y[j]);
				if(s == 0){
					tansaku.push_back(k); 
				}
			}
		int c = tansaku.size(); 
		for(int p = 0; p < c; p++){
			for(int q = p+1; q < c; q++){
				B[tansaku.at(p)][tansaku.at(q)] = true; 
			}
		}
          	ans = (ans + P - A[c]) % P; 
		}
	}
	cout << ans << endl; 
}
```