#include <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const int N = 2000+10;
const int mod = 998244353;
int x[N],y[N];
ll f[N];
 
int cal(int i ,int j,int k)
{
    return (x[j] - x[i])*(y[k] - y[j]) == (y[j] - y[i])*(x[k] - x[j]);
}
int main()
{
    for(int i = 0;i < N;i++) f[i] = i == 0?1:(f[i-1]<<1)%mod;
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n ;i++)
        scanf("%d%d",&x[i],&y[i]);
    ll ans = f[n] - n - 1;
    for(int i = 1;i <= n;i++)
        for(int j = i+1;j <= n;j++)
        {
            int tot = 0;
            for(int k = j+1;k <= n;k++)
                if(cal(i,j,k))  tot++; // 共线
            ans = (ans - f[tot] + mod )%mod;
        }
    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 <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const int N = 2000+10;
const int mod = 998244353;
int x[N],y[N];
ll f[N];
 
int cal(int i ,int j,int k)
{
    return (x[j] - x[i])*(y[k] - y[j]) == (y[j] - y[i])*(x[k] - x[j]);
}
int main()
{
    for(int i = 0;i < N;i++) f[i] = i == 0?1:(f[i-1]<<1)%mod;
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n ;i++)
        scanf("%d%d",&x[i],&y[i]);
    ll ans = f[n] - n - 1;
    for(int i = 1;i <= n;i++)
        for(int j = i+1;j <= n;j++)
        {
            int tot = 0;
            for(int k = j+1;k <= n;k++)
                if(cal(i,j,k))  tot++; // 共线
            ans = (ans - f[tot] + mod )%mod;
        }
    printf("%lld\n",ans);
    return 0;
}
```