#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pb push_back
struct no{
	ll x,y;
	no(){x=0,y=0;}
	no(ll x_,ll y_){x=x_,y=y_;}
	inline db len(){return sqrt(x*x+y*y);}
	inline bool hf(){return x>0||x==0&&y>0;}
};
no operator - (const no &a,const no &b){
	return no(a.x-b.x,a.y-b.y);
}
no operator + (const no &a,const no &b){
	return no(a.x+b.x,a.y+b.y);
}
ll cro(no a,no b){return a.x*b.y-a.y*b.x;}
ll dot(no a,no b){return a.x*b.x+a.y*b.y;}
no st[1001000];int t1,t2,ip[1010000];
db ans;
vector<no>B;
bool vis[1001000];
void mk(vector<no>e,bool cg){
	sort(e.begin(),e.end(),[&](no a,no b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;});
	t1=t2=0;
	int Z=0;
	for(no z:e){
		Z++;
		if(t1&&st[t1].x==z.x)continue;
		while(t1>1&&cro(st[t1]-st[t1-1],z-st[t1-1])<=0)t1--;
		st[++t1]=z,ip[t1]=Z;
	}
	t2=t1;
	Z=0;
	for(no z:e){
		Z++;
		if(t2>t1&&st[t2].x==z.x)t2--;
		while(t2-t1>=2&&cro(st[t2]-st[t2-1],z-st[t2-1])>=0)t2--;
		st[++t2]=z,ip[t2]=Z;
	}
	int nn=e.size();
	for(int i=1;i<=nn;i++)vis[i]=0;
	for(int i=1;i<=t2;i++)vis[ip[i]]=1;
	B.clear();
	if(cg){for(int i=0;i<nn;i++)if(!vis[i+1])B.pb(e[i]);}
	else{for(int i=0;i<nn;i++)if(!vis[i+1])t1=t2=-1;}
}
db calc(){
	int ck=t1-1+(t2-t1-1);
	if(st[1].y!=st[t1+1].y)ck++;
	if(st[t1].y!=st[t2].y)ck++;
	if(ck<3)return -1;
	db su=0;
	su+=(st[t1+1]-st[1]).len();
	su+=(st[t2]-st[t1]).len();
	for(int i=1;i<t1;i++)su+=(st[i+1]-st[i]).len();
	for(int i=t1+1;i<t2;i++)su+=(st[i+1]-st[i]).len();
	return su;
}
bool ck(int L,int R,ll x,ll y){
	if(L==R)return 0;
	no a=no(x,y);
	int l=L,r=R-1,zp=-1;while(l<=r){
		int mi=(l+r)>>1;
		if(x>=st[mi].x)zp=mi,l=mi+1;else r=mi-1;
	}
	assert(zp!=-1);
	return cro(st[zp]-a,st[zp+1]-a)==0;
}
bool isi(ll x,ll y){
	if(x<st[1].x||x>st[t1].x)return 0;
	if(x==st[1].x&&st[1].y<=y&&y<=st[t1+1].y)return 1;
	if(x==st[t1].x&&st[t1].y<=y&&y<=st[t2].y)return 1;
	return ck(1,t1,x,y)||ck(t1+1,t2,x,y);
}
bool chk(no a,no b,no c){return cro(b-a,c-a)>0;}
int n;no a[1010000];
void sol1(){
	vector<no>A;A.pb(no(0,0));
	for(int i=1;i<=n;i++)A.pb(a[i]);
	mk(A,1);
	db sA=calc();if(sA<0)return;
	for(no t:B){
		if(t.x==0&&t.y==0)return;
		if(isi(t.x,t.y))return;
	}
	B.pb(no(0,0));
	mk(B,0);if(t1==-1)return;
	db sB=calc();if(sB<0)return;
	ans=sA+sB;
}
no b[1001000];
db vl[1001000],f[1010000][20];
int L[1010000],R[1010000];
int lg[1010000];
db ask(int l,int r){
	int k=lg[r-l+1];
	return max(f[l][k],f[r-(1<<k)+1][k]);
}
void sol2(){
	sort(a+1,a+n+1,[&](no a,no b){if(a.hf()!=b.hf())return a.hf()<b.hf();return cro(a,b)>0;});
	for(int i=1;i<=n;i++)b[i]=b[i+n]=a[i];
	db os=0;
	for(int i=1;i<=n;i++){
		if(cro(b[i],b[i+1])==0&&dot(b[i],b[i+1])>0)return;
		os+=(b[i+1]-b[i]).len();
	}
	for(int i=1;i<n*2;i++)vl[i]=-(b[i+1]-b[i]).len()+b[i].len()+b[i+1].len(),f[i][0]=vl[i];
	for(int j=1;j<20;j++)for(int i=1;i+(1<<j)<=n*2;i++)
	f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]);
	for(int i=1,j=1;i<=n*2;i++){
		j=max(j,i);
		while(j<n*2&&cro(b[i],b[j+1])>0&&(j==i||chk(b[j-1],b[j],b[j+1])))j++;
		R[i]=j;
	}
	for(int i=n*2,j=n*2;j;i--){
		j=min(j,i);
		while(j>1&&cro(b[j-1],b[i])>0&&(j==i||chk(b[j-1],b[j],b[j+1])))j--;
		L[i]=j;
	}
	for(int i=1;i<=n;i++){
		int l=max(L[n+i]-1,i+2),r=min(n+i-2,R[i+1]);
		if(l>r)continue;
		ans=max(ans,vl[i]+ask(l,r)+os);
		//断掉边(i,i+1),则选择i+1<=j<n+i,j<=R[i+1],L[n+i]<=j+1 
	}
}
void sol(){
	ans=0;
	scanf("%d",&n);ll X,Y;
	for(int i=1;i<=n;i++)scanf("%lld%lld",&X,&Y),a[i]=no(X,Y);
	sol1(),sol2();printf("%.12lf\n",ans);
}
int main(){
	lg[0]=-1;
	for(int i=1;i<=1000000;i++)lg[i]=lg[i>>1]+1;
	int T;scanf("%d",&T);while(T--)sol();
	return 0;
}