#include <stdio.h>
#define MAX_N 201
#define MOD 998244353

int x[MAX_N], y[MAX_N], p[MAX_N];

int main(void)
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d%d", &x[i], &y[i]);
    p[0] = 1;
    for (int i = 0; i < n; i++)
        p[i + 1] = (p[i] << 1) % MOD;
    int ans = p[n] - n - 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < i; j++) {
            int cnt = 0;
            for (int k = 0; k < j; k++) {
                if ((x[i] - x[j]) * (y[i] - y[k]) == (x[i] - x[k]) * (y[i] - y[j]))
                    cnt++;
            }
            ans -= p[cnt];
            ans += MOD;
            ans %= MOD;
        }
    }
    printf("%d\n", ans);
    return 0;
}
### Prompt

Please create a solution in CPP 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 <stdio.h>
#define MAX_N 201
#define MOD 998244353

int x[MAX_N], y[MAX_N], p[MAX_N];

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