//二分图之间只连了2*n条边... 这个思路再想想就有了
#include<bits/stdc++.h>
using namespace std;
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define P pair<int,int>
#define ll long long
#define vi vector<int>
const int N=1e5+5; //不如直接4e5
unordered_map<int,int>mx,my;
vector<P>tu[N*4],eg[N*4];
P pt[N*2],res2[N*2];
int dep[N*4],res1[N*2],cnt1,cnt2;
int dfs(int now,int fa){
    dep[now]=dep[fa]+1; vector<int>q;
    for(auto [son,id]:tu[now]){
        if(!dep[son]){
            int res=dfs(son,now);
            if(res)res2[++cnt2]={res,id};
            else q.push_back(id);
        }
        else if(dep[son]>dep[now])q.push_back(id);
    }
    for(int i=0;i+1<q.size();i+=2)res2[++cnt2]={q[i],q[i+1]};
    if(q.size()&1)return q.back();
    else return 0;
}
void slv(){
    int n;cin>>n;
    mx.clear(); my.clear(); cnt1=cnt2=0;
    rep(i,1,n*2){
        int x,y;cin>>x>>y; pt[i]={x,y};
        mx[x]=1; my[y]=1;
    }
    int cnt=0;
    for(auto &[k,v]:mx)v=++cnt;
    for(auto &[k,v]:my)v=++cnt;
    rep(i,0,cnt){
        tu[i].clear(); eg[i].clear(); dep[i]=0;
    }
    rep(i,1,n*2){
        auto [x,y]=pt[i]; x=mx[x]; y=my[y];
        tu[x].push_back({y,i}); tu[y].push_back({x,i});
    }
    rep(i,1,cnt){
        if(!dep[i]){
            int res=dfs(i,0);
            if(res)res1[++cnt1]=res;
        }
    }
    cout<<cnt2<<'\n'; assert(cnt1%2==0);
    rep(i,1,cnt2)cout<<res2[i].first<<" "<<res2[i].second<<'\n';
    for(int i=1;i<cnt1;i+=2)cout<<res1[i]<<' '<<res1[i+1]<<'\n';
}
signed main(){
    IOS
    int T;cin>>T;
    while(T--)slv();
}