#include <set>
#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define PII pair<int,int>
#define FOR(i,a,b) for (int i=(a);i<=(b);++i)
#define ROF(i,b,a) for (int i=(b);i>=(a);--i)
typedef long long LL;
inline int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}

const int MAXN=300005;
const double eps=1e-8,INF=1e20,PI=3.141592653589793238462;
struct list{
	int l,lt[MAXN],nt[MAXN],x[MAXN];
	void clear(){ l=0; memset(lt,0,sizeof(lt)); }
	void addedge(int a,int b){ l++; nt[l]=lt[a]; lt[a]=l; x[l]=b; }
}E;
struct P{
	double x,y;
	P(double xx=0,double yy=0):x(xx),y(yy){}
	double len()const{ return sqrt(x*x+y*y); }
	P operator +(const P &E)const{ return P(x+E.x,y+E.y); }
	P operator -(const P &E)const{ return P(x-E.x,y-E.y); }
	P operator *(const double &f)const{ return P(x*f,y*f); }
	P operator /(const double &f)const{ return P(x/f,y/f); }
	double operator *(const P &E)const{ return x*E.y-E.x*y; }
	double operator ^(const P &E)const{ return x*E.x+y*E.y; }
	bool operator <(const P &E)const{ return (x<E.x)||(x==E.x&&y<E.y); }
}A[MAXN],B[MAXN];
double arctan(P a){ double ret=atan2(a.y,a.x); return (ret<eps ? ret+PI*2:ret); }
inline double dist(P a,P b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }
inline double dot(P a,P b,P c){ return (a-c)^(b-c); }
inline double cross(P a,P b,P c){ return (a-c)*(b-c); }
struct line{
	double a,b,c;
	line(double aa=0,double bb=0,double cc=0):a(aa),b(bb),c(cc){}
};
inline line get_line(P u,P v){ return line(v.y-u.y,u.x-v.x,v*u); }
P Cramer(line l1,line l2){
	double det=l1.a*l2.b-l1.b*l2.a;
	if (fabs(det)<=eps) return P(INF,INF);
	return P((-l1.c*l2.b+l1.b*l2.c)/det,(-l1.a*l2.c+l1.c*l2.a)/det);
}
double ans;
int n,m,l,a[MAXN],b[MAXN];
int cmp(int x,int y){ return B[x]<B[y]; }
int main(){
	n=read(); E.clear(); l=0;
	FOR(i,1,n) { A[i].x=read(); A[i].y=read(); }
	FOR(i,1,n-1) {
		m=0;
		FOR(j,1,n-1)
		if (j!=i) {
			P D=Cramer(get_line(A[i],A[i+1]),get_line(A[j],A[j+1]));
			if ((D.x-A[i].x)*(D.x-A[i+1].x)>eps||(D.y-A[i].y)*(D.y-A[i+1].y)>eps) continue;
			if ((D.x-A[j].x)*(D.x-A[j+1].x)>eps||(D.y-A[j].y)*(D.y-A[j+1].y)>eps) continue;
			B[a[++m]=++l]=D;
			FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
			//printf("%d %d %d\n",i,j,a[m]);
		}
		B[a[++m]=++l]=A[i];
		FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
		B[a[++m]=++l]=A[i+1];
		FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
		
		sort(a+1,a+1+m,cmp);
		FOR(i,1,m-1)
		if (dist(B[a[i]],B[a[i+1]])>eps) E.addedge(a[i],a[i+1]),E.addedge(a[i+1],a[i]);
	}
	//FOR(i,1,l) printf("%.3lf %.3lf %d\n",B[i].x,B[i].y,i);
	/*FOR(i,1,l) {
		printf("%d:",i);
		for (int k=E.lt[i];k;k=E.nt[k]) printf("%d ",E.x[k]);
		printf("\n");
	}*/
	int x=1,y,z,s;
	double alpha=0,tmp=0;
	FOR(i,2,l) if (B[i]<B[x]) x=i;
	y=E.x[E.lt[x]]; z=s=x;
	for (int i=E.lt[x];i;i=E.nt[i]) {
		int j=E.x[i];
		if (cross(B[j],B[y],B[x])>eps) y=j;
	}
	ans=0;
	while (1) {
		//printf("%d %d %.10lf\n",x,y,alpha);
		ans+=B[x]*B[y];
		if (y==s) break;
		z=x; x=y; y=E.x[E.lt[x]];
		alpha=arctan(B[y]-B[x])-arctan(B[z]-B[x]);
		if (alpha<=eps) alpha+=PI*2;
		for (int i=E.lt[x];i;i=E.nt[i]) {
			int j=E.x[i];
			tmp=arctan(B[j]-B[x])-arctan(B[z]-B[x]);
			if (tmp<=eps) tmp+=PI*2;
			//printf("%d %.10lf %.10lf %.10lf %.5lf %.10lf %.10lf %.5lf\n",j,tmp,(B[j]-B[x]).x,(B[j]-B[x]).y,arctan((B[j]-B[x])),(B[z]-B[x]).x,(B[z]-B[x]).y,arctan((B[z]-B[x])));
			if (tmp<alpha) {
				alpha=tmp;
				y=j;
			}
		}
	}
	printf("%.10lf\n",fabs(ans)/2);
	
	return 0;
}

/*
5
1 1
2 3
3 4
2 5
4 2

*/
### Prompt

Generate a cpp solution to the following problem:

You are given one polygonal line, which is a collection of line segments. Your task is to calculate the sum of areas enclosed by the polygonal line.

A point is defined to be "enclosed" if and only if the point is unreachable without crossing at least one line segment from the point at infinity.



Input

The first line contains one integers N (2 ≤ N ≤ 100). N is the number of segments.

Each of the following N lines consists of two integers Xi and Yi (-105 ≤ Xi, Yi ≤ 105, 1 ≤ i ≤ N) which represents a vertex. A polygonal line is the segments which connect (Xj, Yj) and (Xj+1, Yj+1) ((Xj, Yj) ≠ (Xj+1, Yj+1), 1 ≤ j ≤ N-1). The distance between a segment Sj and all vertices except the end points on segment Sj is guaranteed to be greater than 0.01.

Output

Output the answer in a line. The answer may be printed with an arbitrary number of decimal digits, but may not contain an absolute or relative error greater than or equal to 10-6.

Examples

Input

5
0 0
1 1
1 0
0 1
0 0


Output

0.5


Input

5
0 0
1 1
1 0
0 1
1 0


Output

0.25


Input

21
1 1
-1 1
-1 2
-2 2
-2 1
-1 1
-1 -1
-2 -1
-2 -2
-1 -2
-1 -1
1 -1
1 -2
2 -2
2 -1
1 -1
1 1
2 1
2 2
1 2
1 1


Output

8


Input

16
0 0
1 0
1 1
0 1
0 2
0 3
1 3
1 2
2 2
2 3
3 3
3 2
3 1
2 1
2 0
3 0


Output

0


Input

7
26 52
33 12
-51 68
16 61
43 -26
87 24
12 10


Output

2714.840579710

### Response

```cpp
#include <set>
#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define PII pair<int,int>
#define FOR(i,a,b) for (int i=(a);i<=(b);++i)
#define ROF(i,b,a) for (int i=(b);i>=(a);--i)
typedef long long LL;
inline int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}

const int MAXN=300005;
const double eps=1e-8,INF=1e20,PI=3.141592653589793238462;
struct list{
	int l,lt[MAXN],nt[MAXN],x[MAXN];
	void clear(){ l=0; memset(lt,0,sizeof(lt)); }
	void addedge(int a,int b){ l++; nt[l]=lt[a]; lt[a]=l; x[l]=b; }
}E;
struct P{
	double x,y;
	P(double xx=0,double yy=0):x(xx),y(yy){}
	double len()const{ return sqrt(x*x+y*y); }
	P operator +(const P &E)const{ return P(x+E.x,y+E.y); }
	P operator -(const P &E)const{ return P(x-E.x,y-E.y); }
	P operator *(const double &f)const{ return P(x*f,y*f); }
	P operator /(const double &f)const{ return P(x/f,y/f); }
	double operator *(const P &E)const{ return x*E.y-E.x*y; }
	double operator ^(const P &E)const{ return x*E.x+y*E.y; }
	bool operator <(const P &E)const{ return (x<E.x)||(x==E.x&&y<E.y); }
}A[MAXN],B[MAXN];
double arctan(P a){ double ret=atan2(a.y,a.x); return (ret<eps ? ret+PI*2:ret); }
inline double dist(P a,P b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }
inline double dot(P a,P b,P c){ return (a-c)^(b-c); }
inline double cross(P a,P b,P c){ return (a-c)*(b-c); }
struct line{
	double a,b,c;
	line(double aa=0,double bb=0,double cc=0):a(aa),b(bb),c(cc){}
};
inline line get_line(P u,P v){ return line(v.y-u.y,u.x-v.x,v*u); }
P Cramer(line l1,line l2){
	double det=l1.a*l2.b-l1.b*l2.a;
	if (fabs(det)<=eps) return P(INF,INF);
	return P((-l1.c*l2.b+l1.b*l2.c)/det,(-l1.a*l2.c+l1.c*l2.a)/det);
}
double ans;
int n,m,l,a[MAXN],b[MAXN];
int cmp(int x,int y){ return B[x]<B[y]; }
int main(){
	n=read(); E.clear(); l=0;
	FOR(i,1,n) { A[i].x=read(); A[i].y=read(); }
	FOR(i,1,n-1) {
		m=0;
		FOR(j,1,n-1)
		if (j!=i) {
			P D=Cramer(get_line(A[i],A[i+1]),get_line(A[j],A[j+1]));
			if ((D.x-A[i].x)*(D.x-A[i+1].x)>eps||(D.y-A[i].y)*(D.y-A[i+1].y)>eps) continue;
			if ((D.x-A[j].x)*(D.x-A[j+1].x)>eps||(D.y-A[j].y)*(D.y-A[j+1].y)>eps) continue;
			B[a[++m]=++l]=D;
			FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
			//printf("%d %d %d\n",i,j,a[m]);
		}
		B[a[++m]=++l]=A[i];
		FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
		B[a[++m]=++l]=A[i+1];
		FOR(k,1,l-1) if (dist(B[k],B[l])<=eps) { l--; a[m]=k; break; }
		
		sort(a+1,a+1+m,cmp);
		FOR(i,1,m-1)
		if (dist(B[a[i]],B[a[i+1]])>eps) E.addedge(a[i],a[i+1]),E.addedge(a[i+1],a[i]);
	}
	//FOR(i,1,l) printf("%.3lf %.3lf %d\n",B[i].x,B[i].y,i);
	/*FOR(i,1,l) {
		printf("%d:",i);
		for (int k=E.lt[i];k;k=E.nt[k]) printf("%d ",E.x[k]);
		printf("\n");
	}*/
	int x=1,y,z,s;
	double alpha=0,tmp=0;
	FOR(i,2,l) if (B[i]<B[x]) x=i;
	y=E.x[E.lt[x]]; z=s=x;
	for (int i=E.lt[x];i;i=E.nt[i]) {
		int j=E.x[i];
		if (cross(B[j],B[y],B[x])>eps) y=j;
	}
	ans=0;
	while (1) {
		//printf("%d %d %.10lf\n",x,y,alpha);
		ans+=B[x]*B[y];
		if (y==s) break;
		z=x; x=y; y=E.x[E.lt[x]];
		alpha=arctan(B[y]-B[x])-arctan(B[z]-B[x]);
		if (alpha<=eps) alpha+=PI*2;
		for (int i=E.lt[x];i;i=E.nt[i]) {
			int j=E.x[i];
			tmp=arctan(B[j]-B[x])-arctan(B[z]-B[x]);
			if (tmp<=eps) tmp+=PI*2;
			//printf("%d %.10lf %.10lf %.10lf %.5lf %.10lf %.10lf %.5lf\n",j,tmp,(B[j]-B[x]).x,(B[j]-B[x]).y,arctan((B[j]-B[x])),(B[z]-B[x]).x,(B[z]-B[x]).y,arctan((B[z]-B[x])));
			if (tmp<alpha) {
				alpha=tmp;
				y=j;
			}
		}
	}
	printf("%.10lf\n",fabs(ans)/2);
	
	return 0;
}

/*
5
1 1
2 3
3 4
2 5
4 2

*/
```