#include <cstdio>
#include <cmath>
#include <algorithm>
#define MOD 998244353
using namespace std;
pair<int,int> P[205];
double tmp[205];
int pw2[205];
int N;
int rez;
void fix(int &val)
{
    if(val<0)val+=MOD;
    if(val>=MOD)val-=MOD;
}
void baga(pair<int,int> a)
{
    int len=0;
    for(int i=1;i<=N;i++)
    {

        if(a<P[i])
        {
            tmp[++len]=atan2(P[i].second-a.second,P[i].first-a.first);
        }
    }
    sort(tmp+1,tmp+1+len);
    int cnt=(len>0);
    //printf("%d ",rez);
    fix(rez=rez-1);
    for(int i=2;i<=len;i++)
    {
        if(tmp[i]==tmp[i-1])
        {
            cnt++;
        }
        else
        {
            fix(rez-=(pw2[cnt]-1));
            cnt=1;
        }
    }
    fix(rez-=(pw2[cnt]-1));
//    printf("%d\n",rez);
}
int main()
{
    pw2[0]=1;
    for(int i=1;i<=200;i++)fix(pw2[i]=2*pw2[i-1]);
    scanf("%d",&N);rez=pw2[N]-1;
    for(int i=1;i<=N;i++)scanf("%d %d",&P[i].first,&P[i].second);
    for(int i=1;i<=N;i++)
    {
        baga(P[i]);
    }
    printf("%d",rez);
    return 0;
}

### Prompt

Please formulate a cpp solution 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 <cstdio>
#include <cmath>
#include <algorithm>
#define MOD 998244353
using namespace std;
pair<int,int> P[205];
double tmp[205];
int pw2[205];
int N;
int rez;
void fix(int &val)
{
    if(val<0)val+=MOD;
    if(val>=MOD)val-=MOD;
}
void baga(pair<int,int> a)
{
    int len=0;
    for(int i=1;i<=N;i++)
    {

        if(a<P[i])
        {
            tmp[++len]=atan2(P[i].second-a.second,P[i].first-a.first);
        }
    }
    sort(tmp+1,tmp+1+len);
    int cnt=(len>0);
    //printf("%d ",rez);
    fix(rez=rez-1);
    for(int i=2;i<=len;i++)
    {
        if(tmp[i]==tmp[i-1])
        {
            cnt++;
        }
        else
        {
            fix(rez-=(pw2[cnt]-1));
            cnt=1;
        }
    }
    fix(rez-=(pw2[cnt]-1));
//    printf("%d\n",rez);
}
int main()
{
    pw2[0]=1;
    for(int i=1;i<=200;i++)fix(pw2[i]=2*pw2[i-1]);
    scanf("%d",&N);rez=pw2[N]-1;
    for(int i=1;i<=N;i++)scanf("%d %d",&P[i].first,&P[i].second);
    for(int i=1;i<=N;i++)
    {
        baga(P[i]);
    }
    printf("%d",rez);
    return 0;
}

```