#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
const int N = 210, P = 998244353;
int n,x[N],y[N],po[N];

int main() {
	po[0]=1;
	REP(i,1,N-1) po[i]=po[i-1]*2ll%P;
	scanf("%d",&n);
	REP(i,1,n) scanf("%d%d",x+i,y+i);
	int ans = (po[n]-1-n*(n+1)/2)%P;
	REP(i,1,n) REP(j,i+1,n) {
		int cnt = 0;
		REP(k,j+1,n) if ((x[j]-x[i])*(y[k]-y[i])==(x[k]-x[i])*(y[j]-y[i])) ++cnt;
		ans = (ans-po[cnt]+1)%P;
	}
	if (ans<0) ans+=P;
	printf("%d\n", ans);
}







### Prompt

In cpp, your task is to solve 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 <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
const int N = 210, P = 998244353;
int n,x[N],y[N],po[N];

int main() {
	po[0]=1;
	REP(i,1,N-1) po[i]=po[i-1]*2ll%P;
	scanf("%d",&n);
	REP(i,1,n) scanf("%d%d",x+i,y+i);
	int ans = (po[n]-1-n*(n+1)/2)%P;
	REP(i,1,n) REP(j,i+1,n) {
		int cnt = 0;
		REP(k,j+1,n) if ((x[j]-x[i])*(y[k]-y[i])==(x[k]-x[i])*(y[j]-y[i])) ++cnt;
		ans = (ans-po[cnt]+1)%P;
	}
	if (ans<0) ans+=P;
	printf("%d\n", ans);
}







```