#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const int MaxN = 200;
const int M = 998244353;

int n;
int prePow[MaxN + 1];

struct point
{
	int x, y;
	point() {}
	point(const int &_x, const int &_y)
		: x(_x), y(_y) {}

	friend inline point operator-(const point &lhs, const point &rhs)
	{
		return point(lhs.x - rhs.x, lhs.y - rhs.y);
	}

	friend inline int operator*(const point &lhs, const point &rhs)
	{
		return lhs.x * rhs.y - lhs.y * rhs.x;
	}
};
point po[MaxN + 1];

int main()
{
	cin >> n;
	for (int i = 1; i <= n; ++i)
		cin >> po[i].x >> po[i].y;

	prePow[0] = 1;
	for (int i = 1; i <= n; ++i)
		prePow[i] = prePow[i - 1] * 2 % M;

	int res = (prePow[n] - n - 1 + M) % M;
	for (int i = 1; i < n; ++i)
		for (int j = i + 1; j <= n; ++j)
		{
			int num = 0;
			for (int k = j + 1; k <= n; ++k)
				num += (po[k] - po[i]) * (po[j] - po[i]) == 0;

			res = (res + M - prePow[num]) % M;
		}

	cout << res << endl;

	return 0;
}
### Prompt

In CPP, your task is to solve 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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

const int MaxN = 200;
const int M = 998244353;

int n;
int prePow[MaxN + 1];

struct point
{
	int x, y;
	point() {}
	point(const int &_x, const int &_y)
		: x(_x), y(_y) {}

	friend inline point operator-(const point &lhs, const point &rhs)
	{
		return point(lhs.x - rhs.x, lhs.y - rhs.y);
	}

	friend inline int operator*(const point &lhs, const point &rhs)
	{
		return lhs.x * rhs.y - lhs.y * rhs.x;
	}
};
point po[MaxN + 1];

int main()
{
	cin >> n;
	for (int i = 1; i <= n; ++i)
		cin >> po[i].x >> po[i].y;

	prePow[0] = 1;
	for (int i = 1; i <= n; ++i)
		prePow[i] = prePow[i - 1] * 2 % M;

	int res = (prePow[n] - n - 1 + M) % M;
	for (int i = 1; i < n; ++i)
		for (int j = i + 1; j <= n; ++j)
		{
			int num = 0;
			for (int k = j + 1; k <= n; ++k)
				num += (po[k] - po[i]) * (po[j] - po[i]) == 0;

			res = (res + M - prePow[num]) % M;
		}

	cout << res << endl;

	return 0;
}
```