#include<bits/stdc++.h>
using namespace std;
typedef long double LD;
const int maxn=10010;
const int M=1e6+5;
const LD eps=1e-9;
int sgn(LD x){return (x>eps)-(x<-eps);}

struct P{
	LD x,y;
	int id;
	P(){}
	P(LD _x,LD _y):x(_x),y(_y){}
	LD arg(){return atan2(y,x);}
	LD len(){return sqrt(x*x+y*y);}
	void read(){
		double tx,ty;
		scanf("%lf%lf",&tx,&ty);
		x=tx;y=ty;
	}
	
	void print(){
		printf("(%.3f,%.3f)\n",x,y);
	}
}p[maxn];


P operator+(P a,P b){
	return P(a.x+b.x,a.y+b.y);
}
P operator-(P a,P b){
	return P(a.x-b.x,a.y-b.y);
}
P operator*(P a,LD b){
	return P(a.x*b,a.y*b);
}
P operator/(P a,LD b){
	return P(a.x/b,a.y/b);
}
LD operator^(P a,P b){
	return a.x*b.x+a.y*b.y;
}
LD operator*(P a,P b){
	return a.x*b.y-a.y*b.x;
}
LD det(P a,P b,P c){
	return (b-a)*(c-a);
}

LD dis(P a,P b){
	return (b-a).len();
}
P st;
bool cmp(P a,P b){
	return dis(st,a)<dis(st,b);
}

bool operator<(P a,P b){
	return sgn(a.x-b.x) ? sgn(a.x-b.x)<0 : sgn(a.y-b.y)<0;
}
bool operator==(P a,P b){
	return sgn(a.x-b.x)==0 and sgn(a.y-b.y)==0;
}
int n;


struct L{
	P a,b;
	L(){}
	L(P a,P b):a(a),b(b){}
};
bool onLine(P p,L l){
	return sgn((l.a-p)*(l.b-p))==0;
}
bool onSeg(P p,L s){
	return onLine(p,s) and sgn((p-s.a)^(p-s.b))<=0;
}
bool parallel(L l1,L l2){
	return sgn((l1.a-l1.b)*(l2.a-l2.b))==0;
}
int dir(P p,L l){
	int t=sgn((p-l.b)*(l.b-l.a));
	if(t<0)return -1;
	if(t>0)return 1;
	return 0;
}
P intersect(L l1,L l2){
	LD s1=det(l1.a,l1.b,l2.a);
	LD s2=det(l1.a,l1.b,l2.b);
	return (l2.a*s2-l2.b*s1)/(s2-s1);
}
bool segIntersect(L l1,L l2){
	if(dir(l2.a,l1)*dir(l2.b,l1)<0 && dir(l1.a,l2)*dir(l1.b,l2)<0)
		return true;
	return false;
}
struct edgelist{
	int other[M],succ[M],last[M],sum;
	void clear(){
		memset(last,-1,sizeof last);
		sum=0;
	}
	set<pair<int,int> >S;
	void addEdge(int a,int b){
		if(a>b)swap(a,b);
		if(S.count(make_pair(a,b)))return ;
		S.insert(make_pair(a,b));
	//cerr<<a<<" "<<b<<endl;
		other[sum]=b;succ[sum]=last[a];last[a]=sum++;
		other[sum]=a;succ[sum]=last[b];last[b]=sum++;
	}
}e;

vector<pair<int,LD> >vecs;
vector<int>ee[M];
vector<pair<LD,pair<int,int> > >edges;
int nxt[M],vis[M];
int main(){
	scanf("%d",&n);
	
	
	for(int i=1;i<=n;i++){
		p[i].read();
		p[i].id=i;
	}
	int m=n;
	e.clear();
	
	int sz=0;
	static P pt[M];
	
	for(int i=1;i<n;i++){
		pt[++sz]=p[i];
		for(int j=i+1;j<n;j++){
			L l1=L(p[i],p[i+1]);
			L l2=L(p[j],p[j+1]);
			if(!parallel(l1,l2)){
				P tmp=intersect(l1,l2);
				pt[++sz]=tmp;
			}
		}
	}
	sort(pt+1,pt+1+sz);
	sz=unique(pt+1,pt+1+sz)-pt-1;
	for(int j=1;j<=sz;j++)pt[j].id=j;
	
	for(int i=1;i<n;i++){
		vector<P>vec;
		L l=L(p[i],p[i+1]);
		for(int j=1;j<=sz;j++){
			if(onSeg(pt[j],l)){
				vec.push_back(pt[j]);
			}
		}
		if(vec.empty())assert(0);
		st=p[i];
		sort(vec.begin(),vec.end(),cmp);
		for(int j=1;j<vec.size();j++)
			e.addEdge(vec[j-1].id,vec[j].id);
	}
	
	
	
	n=sz;
	
	//cerr<<sz<<endl;
	//for(int i=1;i<=sz;i++)
	///	pt[i].print();
	LD ans=0;
	for(int x=1;x<=n;x++){
		vector<pair<LD,int> >pairs;
		for(int i=e.last[x];~i;i=e.succ[i]){
			int y=e.other[i];
			pairs.push_back(make_pair((pt[y]-pt[x]).arg(),i));
		}
		sort(pairs.begin(),pairs.end());
		for(int i=0;i<pairs.size();i++){
			nxt[pairs[(i+1)%pairs.size()].second^1]=pairs[i].second;
		}
	}
	memset(vis,0,sizeof vis);
	LD tot=0;
	int cnt=0;
	for(int start=0;start<e.sum;start++){
		if(vis[start])continue;
		LD total=0;
		int now=start;
		vecs.clear();
		while(!vis[now]){
			vis[now]=1;
			total+=pt[e.other[now^1]]*pt[e.other[now]];
			now=nxt[now];
		}
		cnt++;
		if(now==start&&total>0){
			ans+=total;
		}
	}
	//cerr<<cnt<<endl;
	
	printf("%.10f\n",(double)ans/2);
	
	return 0;
}
### Prompt

Please formulate 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<bits/stdc++.h>
using namespace std;
typedef long double LD;
const int maxn=10010;
const int M=1e6+5;
const LD eps=1e-9;
int sgn(LD x){return (x>eps)-(x<-eps);}

struct P{
	LD x,y;
	int id;
	P(){}
	P(LD _x,LD _y):x(_x),y(_y){}
	LD arg(){return atan2(y,x);}
	LD len(){return sqrt(x*x+y*y);}
	void read(){
		double tx,ty;
		scanf("%lf%lf",&tx,&ty);
		x=tx;y=ty;
	}
	
	void print(){
		printf("(%.3f,%.3f)\n",x,y);
	}
}p[maxn];


P operator+(P a,P b){
	return P(a.x+b.x,a.y+b.y);
}
P operator-(P a,P b){
	return P(a.x-b.x,a.y-b.y);
}
P operator*(P a,LD b){
	return P(a.x*b,a.y*b);
}
P operator/(P a,LD b){
	return P(a.x/b,a.y/b);
}
LD operator^(P a,P b){
	return a.x*b.x+a.y*b.y;
}
LD operator*(P a,P b){
	return a.x*b.y-a.y*b.x;
}
LD det(P a,P b,P c){
	return (b-a)*(c-a);
}

LD dis(P a,P b){
	return (b-a).len();
}
P st;
bool cmp(P a,P b){
	return dis(st,a)<dis(st,b);
}

bool operator<(P a,P b){
	return sgn(a.x-b.x) ? sgn(a.x-b.x)<0 : sgn(a.y-b.y)<0;
}
bool operator==(P a,P b){
	return sgn(a.x-b.x)==0 and sgn(a.y-b.y)==0;
}
int n;


struct L{
	P a,b;
	L(){}
	L(P a,P b):a(a),b(b){}
};
bool onLine(P p,L l){
	return sgn((l.a-p)*(l.b-p))==0;
}
bool onSeg(P p,L s){
	return onLine(p,s) and sgn((p-s.a)^(p-s.b))<=0;
}
bool parallel(L l1,L l2){
	return sgn((l1.a-l1.b)*(l2.a-l2.b))==0;
}
int dir(P p,L l){
	int t=sgn((p-l.b)*(l.b-l.a));
	if(t<0)return -1;
	if(t>0)return 1;
	return 0;
}
P intersect(L l1,L l2){
	LD s1=det(l1.a,l1.b,l2.a);
	LD s2=det(l1.a,l1.b,l2.b);
	return (l2.a*s2-l2.b*s1)/(s2-s1);
}
bool segIntersect(L l1,L l2){
	if(dir(l2.a,l1)*dir(l2.b,l1)<0 && dir(l1.a,l2)*dir(l1.b,l2)<0)
		return true;
	return false;
}
struct edgelist{
	int other[M],succ[M],last[M],sum;
	void clear(){
		memset(last,-1,sizeof last);
		sum=0;
	}
	set<pair<int,int> >S;
	void addEdge(int a,int b){
		if(a>b)swap(a,b);
		if(S.count(make_pair(a,b)))return ;
		S.insert(make_pair(a,b));
	//cerr<<a<<" "<<b<<endl;
		other[sum]=b;succ[sum]=last[a];last[a]=sum++;
		other[sum]=a;succ[sum]=last[b];last[b]=sum++;
	}
}e;

vector<pair<int,LD> >vecs;
vector<int>ee[M];
vector<pair<LD,pair<int,int> > >edges;
int nxt[M],vis[M];
int main(){
	scanf("%d",&n);
	
	
	for(int i=1;i<=n;i++){
		p[i].read();
		p[i].id=i;
	}
	int m=n;
	e.clear();
	
	int sz=0;
	static P pt[M];
	
	for(int i=1;i<n;i++){
		pt[++sz]=p[i];
		for(int j=i+1;j<n;j++){
			L l1=L(p[i],p[i+1]);
			L l2=L(p[j],p[j+1]);
			if(!parallel(l1,l2)){
				P tmp=intersect(l1,l2);
				pt[++sz]=tmp;
			}
		}
	}
	sort(pt+1,pt+1+sz);
	sz=unique(pt+1,pt+1+sz)-pt-1;
	for(int j=1;j<=sz;j++)pt[j].id=j;
	
	for(int i=1;i<n;i++){
		vector<P>vec;
		L l=L(p[i],p[i+1]);
		for(int j=1;j<=sz;j++){
			if(onSeg(pt[j],l)){
				vec.push_back(pt[j]);
			}
		}
		if(vec.empty())assert(0);
		st=p[i];
		sort(vec.begin(),vec.end(),cmp);
		for(int j=1;j<vec.size();j++)
			e.addEdge(vec[j-1].id,vec[j].id);
	}
	
	
	
	n=sz;
	
	//cerr<<sz<<endl;
	//for(int i=1;i<=sz;i++)
	///	pt[i].print();
	LD ans=0;
	for(int x=1;x<=n;x++){
		vector<pair<LD,int> >pairs;
		for(int i=e.last[x];~i;i=e.succ[i]){
			int y=e.other[i];
			pairs.push_back(make_pair((pt[y]-pt[x]).arg(),i));
		}
		sort(pairs.begin(),pairs.end());
		for(int i=0;i<pairs.size();i++){
			nxt[pairs[(i+1)%pairs.size()].second^1]=pairs[i].second;
		}
	}
	memset(vis,0,sizeof vis);
	LD tot=0;
	int cnt=0;
	for(int start=0;start<e.sum;start++){
		if(vis[start])continue;
		LD total=0;
		int now=start;
		vecs.clear();
		while(!vis[now]){
			vis[now]=1;
			total+=pt[e.other[now^1]]*pt[e.other[now]];
			now=nxt[now];
		}
		cnt++;
		if(now==start&&total>0){
			ans+=total;
		}
	}
	//cerr<<cnt<<endl;
	
	printf("%.10f\n",(double)ans/2);
	
	return 0;
}
```