#include <bits/stdc++.h>

using namespace std;

#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef pair<int, int> pi;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF;

const int MAX_N = 2e2 + 10, MOD = 998244353;

int N; pi Nr[MAX_N];
int F[MAX_N];
ll cross(pi a, pi b, pi c) {
	pi x = pi(b.one - a.one, b.two - a.two);
	pi y = pi(c.one - a.one, c.two - a.two);
	return x.one * y.two - y.one * x.two;
}
int main() {
	cin >> N;
	for(int i=0; i<N; i++) scanf("%d%d", &Nr[i].one, &Nr[i].two);
	
	F[0] = 1; for(int i=1; i<=N; i++) F[i] = 1ll * F[i-1] * 2 % MOD;

	int ans = (F[N] - N - 1) % MOD;
	for(int i=0; i<N; i++) for(int j=i+1; j<N; j++) {
		bool flag = false;
		for(int k=0; k<j; k++) if(k != i && k != j && cross(Nr[i], Nr[j], Nr[k]) == 0) {flag = true; break;}
		if(flag) continue;
		int cnt = 2;
		for(int k=j+1; k<N; k++) if(cross(Nr[i], Nr[j], Nr[k]) == 0) cnt++;
		ans -= (F[cnt] - cnt - 1); ans %= MOD;
	}
	printf("%d\n", (ans + MOD) % MOD);
	return 0;
}

### Prompt

Your challenge is to write 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>

using namespace std;

#define SZ(v) ((int)(v).size())
#define ALL(v) (v).begin(),(v).end()
#define one first
#define two second
typedef long long ll;
typedef pair<int, int> pi;
const int INF = 0x3f2f1f0f;
const ll LINF = 1ll * INF * INF;

const int MAX_N = 2e2 + 10, MOD = 998244353;

int N; pi Nr[MAX_N];
int F[MAX_N];
ll cross(pi a, pi b, pi c) {
	pi x = pi(b.one - a.one, b.two - a.two);
	pi y = pi(c.one - a.one, c.two - a.two);
	return x.one * y.two - y.one * x.two;
}
int main() {
	cin >> N;
	for(int i=0; i<N; i++) scanf("%d%d", &Nr[i].one, &Nr[i].two);
	
	F[0] = 1; for(int i=1; i<=N; i++) F[i] = 1ll * F[i-1] * 2 % MOD;

	int ans = (F[N] - N - 1) % MOD;
	for(int i=0; i<N; i++) for(int j=i+1; j<N; j++) {
		bool flag = false;
		for(int k=0; k<j; k++) if(k != i && k != j && cross(Nr[i], Nr[j], Nr[k]) == 0) {flag = true; break;}
		if(flag) continue;
		int cnt = 2;
		for(int k=j+1; k<N; k++) if(cross(Nr[i], Nr[j], Nr[k]) == 0) cnt++;
		ans -= (F[cnt] - cnt - 1); ans %= MOD;
	}
	printf("%d\n", (ans + MOD) % MOD);
	return 0;
}

```