#include<bits/stdc++.h>
#define rep(i,a,b) for (int i=(a); i<=int(b); i++)
using namespace std;

const int maxn = 205, MOD = 998244353;
int x[maxn], y[maxn], bin[maxn], n, ans, cnt;

inline int mul(int a, int b) {
	return 1LL * a * b % MOD;
}
inline int add(int a, int b) {
	a += b;
	if (a >= MOD) a -= MOD;
	if (a < 0) a += MOD;
	return a;
}

int main() {
	scanf("%d", &n); bin[0] = 1;
	rep (i, 1, n) scanf("%d%d", &x[i], &y[i]);
	rep (i, 1, n) bin[i] = bin[i-1] * 2 % MOD;
	ans = add(bin[n], -n-1);
	rep (i, 1, n) rep (j, i+1, n) {
		cnt = 0;
		rep (k, j+1, n)
			if ((x[i]-x[k])*(y[j]-y[k]) - (x[j]-x[k])*(y[i]-y[k]) == 0) cnt++;
		ans = add(ans, -bin[cnt]);
	}
	ans = (ans % MOD + MOD) % MOD;
	printf("%d\n", ans);
	return 0;
}
### 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>
#define rep(i,a,b) for (int i=(a); i<=int(b); i++)
using namespace std;

const int maxn = 205, MOD = 998244353;
int x[maxn], y[maxn], bin[maxn], n, ans, cnt;

inline int mul(int a, int b) {
	return 1LL * a * b % MOD;
}
inline int add(int a, int b) {
	a += b;
	if (a >= MOD) a -= MOD;
	if (a < 0) a += MOD;
	return a;
}

int main() {
	scanf("%d", &n); bin[0] = 1;
	rep (i, 1, n) scanf("%d%d", &x[i], &y[i]);
	rep (i, 1, n) bin[i] = bin[i-1] * 2 % MOD;
	ans = add(bin[n], -n-1);
	rep (i, 1, n) rep (j, i+1, n) {
		cnt = 0;
		rep (k, j+1, n)
			if ((x[i]-x[k])*(y[j]-y[k]) - (x[j]-x[k])*(y[i]-y[k]) == 0) cnt++;
		ans = add(ans, -bin[cnt]);
	}
	ans = (ans % MOD + MOD) % MOD;
	printf("%d\n", ans);
	return 0;
}
```