#include <bits/stdc++.h>
#define ll long long
#define mk make_pair
using namespace std;
const int N = 2e2 + 5;
const int mod = 998244353;
bool bo[N][N];
int n, x[N], y[N];
vector<int> V;
ll ans, p[N];
bool line(int i, int j, int k) {
	return (x[j] - x[i]) * (y[k] - y[i]) - (x[k] - x[i]) * (y[j] - y[i]) == 0; 
}
int main() {
	p[0] = 1;
	for (int i = 1; i < N; i++) p[i] = p[i - 1] * 2 % mod;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> x[i] >> y[i];
	}
	for (int i = 1; i <= n; i++) {
		for (int j = i + 1; j <= n; j++) {
			if (bo[i][j]) continue;
			V.clear();
			V.push_back(i);
			V.push_back(j);
			for (int k = j + 1; k <= n; k++) {
				if (line(i, j, k)) V.push_back(k);
			}
			for (int k = 0; k < V.size(); k++)
				for (int l = 0; l < V.size(); l++)
					bo[V[k]][V[l]] = 1;
			ans += p[V.size()] - V.size() - 1;
			ans %= mod;
		}
	}
	ans = p[n] - ans - n - 1;
	ans %= mod;
	ans += mod;
	ans %= mod;
	cout << ans;
}


### Prompt

Your task is to create 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>
#define ll long long
#define mk make_pair
using namespace std;
const int N = 2e2 + 5;
const int mod = 998244353;
bool bo[N][N];
int n, x[N], y[N];
vector<int> V;
ll ans, p[N];
bool line(int i, int j, int k) {
	return (x[j] - x[i]) * (y[k] - y[i]) - (x[k] - x[i]) * (y[j] - y[i]) == 0; 
}
int main() {
	p[0] = 1;
	for (int i = 1; i < N; i++) p[i] = p[i - 1] * 2 % mod;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> x[i] >> y[i];
	}
	for (int i = 1; i <= n; i++) {
		for (int j = i + 1; j <= n; j++) {
			if (bo[i][j]) continue;
			V.clear();
			V.push_back(i);
			V.push_back(j);
			for (int k = j + 1; k <= n; k++) {
				if (line(i, j, k)) V.push_back(k);
			}
			for (int k = 0; k < V.size(); k++)
				for (int l = 0; l < V.size(); l++)
					bo[V[k]][V[l]] = 1;
			ans += p[V.size()] - V.size() - 1;
			ans %= mod;
		}
	}
	ans = p[n] - ans - n - 1;
	ans %= mod;
	ans += mod;
	ans %= mod;
	cout << ans;
}


```