#include <cstdio>
#include <iostream>
using namespace std;

typedef long long LL;

const int N = 205;
const LL mod = 998244353LL;

LL ans, base[N];
bool done[N][N];
LL x[N], y[N];
int sta[N], top;

inline LL mul(int a, int b, int c) {
  return (x[b] - x[a]) * (y[c] - y[a]) - (x[c] - x[a]) * (y[b] - y[a]);
}

int main() {
  int n;
  scanf("%d", &n);
  ans = 1;
  base[0] = 1;
  for (int i = 1; i <= n; i ++) base[i] = base[i - 1] * 2LL % mod;
  ans = base[n];
  for (int i = 1; i <= n; i ++) scanf("%lld%lld", &x[i], &y[i]);
  ans = (ans + mod - n - 1) % mod;
  for (int i = 1; i <= n; i ++)
    for (int j = 1; j <= n; j ++)
      if (!done[i][j] && i != j) {
	top = 0;
	sta[++ top] = i;
	sta[++ top] = j;
	for (int k = 1; k <= n; k ++)
	  if (k != i && k != j && mul(i, j, k) == 0) sta[++ top] = k;
	ans = (mod + ans + top + 1 - base[top]) % mod;
	for (int k = 1; k <= top; k ++)
	  for (int l = 1; l < k; l ++) done[sta[k]][sta[l]] = done[sta[l]][sta[k]] = 1;
      }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 <cstdio>
#include <iostream>
using namespace std;

typedef long long LL;

const int N = 205;
const LL mod = 998244353LL;

LL ans, base[N];
bool done[N][N];
LL x[N], y[N];
int sta[N], top;

inline LL mul(int a, int b, int c) {
  return (x[b] - x[a]) * (y[c] - y[a]) - (x[c] - x[a]) * (y[b] - y[a]);
}

int main() {
  int n;
  scanf("%d", &n);
  ans = 1;
  base[0] = 1;
  for (int i = 1; i <= n; i ++) base[i] = base[i - 1] * 2LL % mod;
  ans = base[n];
  for (int i = 1; i <= n; i ++) scanf("%lld%lld", &x[i], &y[i]);
  ans = (ans + mod - n - 1) % mod;
  for (int i = 1; i <= n; i ++)
    for (int j = 1; j <= n; j ++)
      if (!done[i][j] && i != j) {
	top = 0;
	sta[++ top] = i;
	sta[++ top] = j;
	for (int k = 1; k <= n; k ++)
	  if (k != i && k != j && mul(i, j, k) == 0) sta[++ top] = k;
	ans = (mod + ans + top + 1 - base[top]) % mod;
	for (int k = 1; k <= top; k ++)
	  for (int l = 1; l < k; l ++) done[sta[k]][sta[l]] = done[sta[l]][sta[k]] = 1;
      }
  printf("%lld\n", ans);
  return 0;
}

```