#include <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;
const int MAXN = 205;
const int mod = 998244353;

int n, pwr[MAXN];
pi a[MAXN];

int ccw(pi a, pi b, pi c){
	int dx1 = b.first - a.first;
	int dy1 = b.second - a.second;
	int dx2 = c.first - a.first;
	int dy2 = c.second - a.second;
	return dx1 * dy2 - dy1 * dx2;
}

int main(){
	int n; cin >> n;
	pwr[0] = 1;
	for(int i=0; i<n; i++){
		pwr[i+1] = pwr[i] * 2 % mod;
		cin >> a[i].first >> a[i].second;
	}
	sort(a, a + n);
	lint ret = 1;
	for(int i=0; i<n; i++){
		sort(a + i + 1, a + n, [&](const pi &p, const pi &q){
			return ccw(a[i], p, q) > 0;
		});
		ret += 1;
		for(int j=i+1; j<n; ){
			int e = j;
			while(e < n && ccw(a[i], a[j], a[e]) == 0) e++;
		//	printf("%d %d\n", j, e);
			ret += pwr[e - j] - 1;
			j = e;
		}
		sort(a + i + 1, a + n);
	}
	ret %= mod;
	ret = mod - ret + pwr[n];
	ret %= mod;
	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 <bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef pair<int, int> pi;
const int MAXN = 205;
const int mod = 998244353;

int n, pwr[MAXN];
pi a[MAXN];

int ccw(pi a, pi b, pi c){
	int dx1 = b.first - a.first;
	int dy1 = b.second - a.second;
	int dx2 = c.first - a.first;
	int dy2 = c.second - a.second;
	return dx1 * dy2 - dy1 * dx2;
}

int main(){
	int n; cin >> n;
	pwr[0] = 1;
	for(int i=0; i<n; i++){
		pwr[i+1] = pwr[i] * 2 % mod;
		cin >> a[i].first >> a[i].second;
	}
	sort(a, a + n);
	lint ret = 1;
	for(int i=0; i<n; i++){
		sort(a + i + 1, a + n, [&](const pi &p, const pi &q){
			return ccw(a[i], p, q) > 0;
		});
		ret += 1;
		for(int j=i+1; j<n; ){
			int e = j;
			while(e < n && ccw(a[i], a[j], a[e]) == 0) e++;
		//	printf("%d %d\n", j, e);
			ret += pwr[e - j] - 1;
			j = e;
		}
		sort(a + i + 1, a + n);
	}
	ret %= mod;
	ret = mod - ret + pwr[n];
	ret %= mod;
	cout << ret << endl;
}

```