#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define pi pair<int,int>
#define pl pair<long long,long long>
#define lg length()
#define pb push_back
#define MAXN 100005
#define INF 1000000005
#define LINF 1000000000000000005

long long n,x[205],y[205],mod=998244353,cnt,v[205],s,p[205];

int32_t main(){
    ios_base :: sync_with_stdio(0); cin.tie(); cout.tie();
    cin >> n; p[0]=1;
    for(int i=1;i<=n;i++){
        cin >> x[i] >> y[i];
        p[i]=p[i-1]*2%mod;
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++) v[j]=0;
        for(int j=i+1;j<=n;j++){
            if(v[j]) continue;
            v[j]=1;
            s=1;
            for(int t=j+1;t<=n;t++){
                if((x[t]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[t]-y[i])) s++,v[t]=1;
            }
            cnt+=(p[s]-1);
            cnt%=mod;
        }
    }
    cnt=(p[n]-cnt-1-n+2*mod)%mod;
    cout << cnt;
}

### 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>
using namespace std;
#define x first
#define y second
#define dbg(x) cout << #x << '=' << x << '\n';
#define ll long long
#define pi pair<int,int>
#define pl pair<long long,long long>
#define lg length()
#define pb push_back
#define MAXN 100005
#define INF 1000000005
#define LINF 1000000000000000005

long long n,x[205],y[205],mod=998244353,cnt,v[205],s,p[205];

int32_t main(){
    ios_base :: sync_with_stdio(0); cin.tie(); cout.tie();
    cin >> n; p[0]=1;
    for(int i=1;i<=n;i++){
        cin >> x[i] >> y[i];
        p[i]=p[i-1]*2%mod;
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++) v[j]=0;
        for(int j=i+1;j<=n;j++){
            if(v[j]) continue;
            v[j]=1;
            s=1;
            for(int t=j+1;t<=n;t++){
                if((x[t]-x[i])*(y[j]-y[i])==(x[j]-x[i])*(y[t]-y[i])) s++,v[t]=1;
            }
            cnt+=(p[s]-1);
            cnt%=mod;
        }
    }
    cnt=(p[n]-cnt-1-n+2*mod)%mod;
    cout << cnt;
}

```