#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const ll MAX=204;
const ll MOD=998244353;
pll pt[MAX];
ll chk[MAX][MAX],ans,n;
vector<int> tmp;
ll power(ll x,ll y){
    if(y==0) return 1;
    ll t=power(x,y/2);
    if(y%2) return t*t%MOD*x%MOD;
    return t*t%MOD;
}
ll func(ll x){
    return (power(2,x)-x*(x-1)/2-x-1+MOD)%MOD;
}
int main(){
    scanf("%lld",&n);
    ans=func(n);
    for(ll i=0;i<n;i++) scanf("%lld%lld",&pt[i].x,&pt[i].y);
    for(ll i=0;i<n;i++) for(ll j=i+1;j<n;j++) if(chk[i][j]==0){
        ll cnt=2;
        tmp.push_back(i),tmp.push_back(j);
        for(ll k=j+1;k<n;k++) if((pt[k].x-pt[j].x)*(pt[j].y-pt[i].y)==(pt[k].y-pt[j].y)*(pt[j].x-pt[i].x)){
            tmp.push_back(k);
        }
        for(ll k=0;k<tmp.size();k++) for(ll l=0;l<tmp.size();l++) chk[tmp[k]][tmp[l]]=1;
        if(tmp.size()>=3) ans+=MOD-func(tmp.size()),ans%=MOD;
        tmp.clear();
    }
    printf("%lld\n",ans);
}

### Prompt

Your task is to create 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<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
const ll MAX=204;
const ll MOD=998244353;
pll pt[MAX];
ll chk[MAX][MAX],ans,n;
vector<int> tmp;
ll power(ll x,ll y){
    if(y==0) return 1;
    ll t=power(x,y/2);
    if(y%2) return t*t%MOD*x%MOD;
    return t*t%MOD;
}
ll func(ll x){
    return (power(2,x)-x*(x-1)/2-x-1+MOD)%MOD;
}
int main(){
    scanf("%lld",&n);
    ans=func(n);
    for(ll i=0;i<n;i++) scanf("%lld%lld",&pt[i].x,&pt[i].y);
    for(ll i=0;i<n;i++) for(ll j=i+1;j<n;j++) if(chk[i][j]==0){
        ll cnt=2;
        tmp.push_back(i),tmp.push_back(j);
        for(ll k=j+1;k<n;k++) if((pt[k].x-pt[j].x)*(pt[j].y-pt[i].y)==(pt[k].y-pt[j].y)*(pt[j].x-pt[i].x)){
            tmp.push_back(k);
        }
        for(ll k=0;k<tmp.size();k++) for(ll l=0;l<tmp.size();l++) chk[tmp[k]][tmp[l]]=1;
        if(tmp.size()>=3) ans+=MOD-func(tmp.size()),ans%=MOD;
        tmp.clear();
    }
    printf("%lld\n",ans);
}

```