#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)

using namespace std;

typedef long long LL;
const int N = 210;
const int mod = 998244353;

int x[N], y[N];
bool vis[N];

LL qpow(LL a, LL b)
{
	LL ret = 1, base = a;
	while (b) {
		if (b & 1) ret = ret * base % mod;
		base = base * base % mod, b >>= 1;
	}
	return ret;
}

int main()
{
	int n;
	scanf("%d", &n);
	Rep(i, n)scanf("%d%d", &x[i], &y[i]);
	
	LL ans = qpow(2, n);
	Rep(i, n) Rep(j, n) if (i != j){
		int cnt = 2;
		Rep(k, n) if (k != i && k != j)
			if ((y[j] - y[i]) * (x[k] - x[i]) == (y[k] - y[i]) * (x[j] - x[i])) cnt ++;
		LL tmp = (qpow(2, cnt) - 1 - cnt + mod) % mod;
		ans = (ans - tmp * qpow(cnt * (cnt - 1), mod - 2) % mod + mod) % mod;
		//printf("%lld %d\n", tmp, cnt * (cnt - 1));
	}
	
	ans = (ans - n - 1 + mod) % mod;
	printf("%lld\n", ans);
	
	
	return 0;
}

### Prompt

Generate 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 <cstdio>
#include <algorithm>
#include <cstring>
#include <ctime>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)
#define RepG(i, x) for (int i = head[x]; i; i = edge[i].next)
#define v edge[i].to
#define mp(a, b) make_pair(a, b)

using namespace std;

typedef long long LL;
const int N = 210;
const int mod = 998244353;

int x[N], y[N];
bool vis[N];

LL qpow(LL a, LL b)
{
	LL ret = 1, base = a;
	while (b) {
		if (b & 1) ret = ret * base % mod;
		base = base * base % mod, b >>= 1;
	}
	return ret;
}

int main()
{
	int n;
	scanf("%d", &n);
	Rep(i, n)scanf("%d%d", &x[i], &y[i]);
	
	LL ans = qpow(2, n);
	Rep(i, n) Rep(j, n) if (i != j){
		int cnt = 2;
		Rep(k, n) if (k != i && k != j)
			if ((y[j] - y[i]) * (x[k] - x[i]) == (y[k] - y[i]) * (x[j] - x[i])) cnt ++;
		LL tmp = (qpow(2, cnt) - 1 - cnt + mod) % mod;
		ans = (ans - tmp * qpow(cnt * (cnt - 1), mod - 2) % mod + mod) % mod;
		//printf("%lld %d\n", tmp, cnt * (cnt - 1));
	}
	
	ans = (ans - n - 1 + mod) % mod;
	printf("%lld\n", ans);
	
	
	return 0;
}

```