#include<bits/stdc++.h>
#define pi 3.1415926535897932384626
using namespace std;
      
struct P{
  int x,y;
  bool operator < (const P &a )const{
    //if(y!=a.y)
    return y<a.y;
  };
};

int a,b,c,d,i,k,n,m,e,dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1};
long long o[1101];
int l[1111];
int j[1];

long long x,y,z,mod=998244353;
//string r;
char r[1];

P u[1111];
queue<int> q;
stack<P> s;
vector<int> v[1111];

bool as(P a,P b)
{
  return a.x<b.x;
}

int ccw(P a,P b)
{
  return a.x*b.y-a.y*b.x;
}

int main()
{
  scanf("%d",&a);
  for(int t=1;t<=a;t++)
  {
    scanf("%d %d",&n,&m);
    u[t]={n,m};
  }
  o[0]=1;
  for(int t=1;t<=333;t++)
  o[t]=o[t-1]*2%mod;
  x=(o[a]-a-1+mod)%mod;
  for(int t=1;t<a;t++)
    for(int w=t+1;w<=a;w++)
    {
      int k=0;
      for(int h=w+1;h<=a;h++)
        if(ccw({u[t].x-u[h].x,u[t].y-u[h].y},{u[t].x-u[w].x,u[t].y-u[w].y})==0) k++;
      x-=(o[k]+mod)%mod;
      x=(x+mod)%mod;
      //printf("%lld\n",x);
    }
  printf("%lld",x);
}
//3



### Prompt

Create a solution in CPP for 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<bits/stdc++.h>
#define pi 3.1415926535897932384626
using namespace std;
      
struct P{
  int x,y;
  bool operator < (const P &a )const{
    //if(y!=a.y)
    return y<a.y;
  };
};

int a,b,c,d,i,k,n,m,e,dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1};
long long o[1101];
int l[1111];
int j[1];

long long x,y,z,mod=998244353;
//string r;
char r[1];

P u[1111];
queue<int> q;
stack<P> s;
vector<int> v[1111];

bool as(P a,P b)
{
  return a.x<b.x;
}

int ccw(P a,P b)
{
  return a.x*b.y-a.y*b.x;
}

int main()
{
  scanf("%d",&a);
  for(int t=1;t<=a;t++)
  {
    scanf("%d %d",&n,&m);
    u[t]={n,m};
  }
  o[0]=1;
  for(int t=1;t<=333;t++)
  o[t]=o[t-1]*2%mod;
  x=(o[a]-a-1+mod)%mod;
  for(int t=1;t<a;t++)
    for(int w=t+1;w<=a;w++)
    {
      int k=0;
      for(int h=w+1;h<=a;h++)
        if(ccw({u[t].x-u[h].x,u[t].y-u[h].y},{u[t].x-u[w].x,u[t].y-u[w].y})==0) k++;
      x-=(o[k]+mod)%mod;
      x=(x+mod)%mod;
      //printf("%lld\n",x);
    }
  printf("%lld",x);
}
//3



```