#include <bits/stdc++.h>

using namespace std;

const int mod = 998244353;

bool vis[1001][1001];
int a[1001], b[1001];
int stk[1001];
int mul[1001];
int ksm(int x, int y) { return mul[y]; }
int main( ) {
	int n;
	int ans = 0;
	scanf("%d", &n);
	mul[0] = 1;
	for (int i = 1; i <= n; i ++) mul[i] = mul[i - 1] * 2 % mod;
	for (int i = 1; i <= n; i ++) {
		scanf("%d %d", &a[i], &b[i]);
	}
	for (int i = 1; i <= n; i ++)
		for (int j = i + 1; j <= n; j ++) {
			if (vis[i][j]) continue;
			int cnt = 2;
			stk[1] = i, stk[2] = j;
			for (int k = 1; k <= n; k ++) {
				if (k == i || k == j) continue;
				pair<int, int> tmpx = make_pair(a[k] - a[i], b[k] - b[i]),
							   tmpy = make_pair(a[k] - a[j], b[k] - b[j]);
				if (tmpx.first * tmpy.second == tmpx.second * tmpy.first)
					stk[++ cnt] = k;
			}
			for (int k = 1; k <= cnt; k ++)
				for (int l = k + 1; l <= cnt; ++ l)
					vis[stk[k]][stk[l]] = true;
			ans += (0LL + ksm(2, cnt) - (cnt + 1) % mod + mod) % mod;
			ans %= mod;
		}
	printf("%d\n", ((1LL * ksm(2, n) - 1 - n - ans) % mod + mod) % mod);
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;

const int mod = 998244353;

bool vis[1001][1001];
int a[1001], b[1001];
int stk[1001];
int mul[1001];
int ksm(int x, int y) { return mul[y]; }
int main( ) {
	int n;
	int ans = 0;
	scanf("%d", &n);
	mul[0] = 1;
	for (int i = 1; i <= n; i ++) mul[i] = mul[i - 1] * 2 % mod;
	for (int i = 1; i <= n; i ++) {
		scanf("%d %d", &a[i], &b[i]);
	}
	for (int i = 1; i <= n; i ++)
		for (int j = i + 1; j <= n; j ++) {
			if (vis[i][j]) continue;
			int cnt = 2;
			stk[1] = i, stk[2] = j;
			for (int k = 1; k <= n; k ++) {
				if (k == i || k == j) continue;
				pair<int, int> tmpx = make_pair(a[k] - a[i], b[k] - b[i]),
							   tmpy = make_pair(a[k] - a[j], b[k] - b[j]);
				if (tmpx.first * tmpy.second == tmpx.second * tmpy.first)
					stk[++ cnt] = k;
			}
			for (int k = 1; k <= cnt; k ++)
				for (int l = k + 1; l <= cnt; ++ l)
					vis[stk[k]][stk[l]] = true;
			ans += (0LL + ksm(2, cnt) - (cnt + 1) % mod + mod) % mod;
			ans %= mod;
		}
	printf("%d\n", ((1LL * ksm(2, n) - 1 - n - ans) % mod + mod) % mod);
	return 0;
}

```