#include<bits/stdc++.h>
#define Mo 998244353
#define N 500
typedef long long LL;
using namespace std;
int x[N], y[N], b[N], num[N], n;

void prepare() {
    b[0] = 1;
    for(int i = 1; i <= n; i++) b[i] = ((LL)b[i - 1] * 2) % Mo;
    for(int i = 1; i <= n; i++) b[i] = (b[i] - 1 - i - i * (i - 1) / 2) % Mo;
}

int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> x[i] >> y[i];
    prepare();
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++) {
            if (i == j) continue;
            int cnt = 2;
            for(int k = 1; k <= n; k++) {
                if (k == i || k == j) continue;
                if ((LL)(x[k] - x[i]) * (y[k] - y[j]) == (LL)(x[k] - x[j]) * (y[k] - y[i])) cnt++;
            }
            num[cnt]++;
        }
    for(int i = 3; i <= n; i++) num[i] = num[i] / i / (i - 1);
    LL ans = b[n];
    for(int i = 3; i <= n; i++) ans = (ans - (LL)num[i] * b[i] % Mo) % Mo;
    cout << (ans % Mo + Mo) % Mo << endl;
}

### 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>
#define Mo 998244353
#define N 500
typedef long long LL;
using namespace std;
int x[N], y[N], b[N], num[N], n;

void prepare() {
    b[0] = 1;
    for(int i = 1; i <= n; i++) b[i] = ((LL)b[i - 1] * 2) % Mo;
    for(int i = 1; i <= n; i++) b[i] = (b[i] - 1 - i - i * (i - 1) / 2) % Mo;
}

int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> x[i] >> y[i];
    prepare();
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++) {
            if (i == j) continue;
            int cnt = 2;
            for(int k = 1; k <= n; k++) {
                if (k == i || k == j) continue;
                if ((LL)(x[k] - x[i]) * (y[k] - y[j]) == (LL)(x[k] - x[j]) * (y[k] - y[i])) cnt++;
            }
            num[cnt]++;
        }
    for(int i = 3; i <= n; i++) num[i] = num[i] / i / (i - 1);
    LL ans = b[n];
    for(int i = 3; i <= n; i++) ans = (ans - (LL)num[i] * b[i] % Mo) % Mo;
    cout << (ans % Mo + Mo) % Mo << endl;
}

```