Problem: A bitstring is a string that contains only the characters 0 and 1.

Koyomi Kanou is working hard towards her dream of becoming a writer. To practice, she decided to participate in the Binary Novel Writing Contest. The writing prompt for the contest consists of three bitstrings of length 2n. A valid novel for the contest is a bitstring of length at most 3n that contains at least two of the three given strings as subsequences.

Koyomi has just received the three prompt strings from the contest organizers. Help her write a valid novel for the contest.

A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero) characters.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5).

Each of the following three lines contains a bitstring of length 2n. It is guaranteed that these three strings are pairwise distinct.

It is guaranteed that the sum of n across all test cases does not exceed 10^5.

Output

For each test case, print a single line containing a bitstring of length at most 3n that has at least two of the given bitstrings as subsequences.

It can be proven that under the constraints of the problem, such a bitstring always exists.

If there are multiple possible answers, you may output any of them.

Example

Input


2
1
00
11
01
3
011001
111010
010001


Output


010
011001010

Note

In the first test case, the bitstrings 00 and 01 are subsequences of the output string: 010 and 010. Note that 11 is not a subsequence of the output string, but this is not required.

In the second test case all three input strings are subsequences of the output string: 011001010, 011001010 and 011001010.

Solution (in CPP):

Solution:
#pragma GCC target("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
// #include<ext/pb_ds/assoc_container.hpp>
// #include<ext/pb_ds/tree_policy.hpp>
// #include<ext/pb_ds/tag_and_trait.hpp>
// using namespace __gnu_pbds;
// #include<boost/multiprecision/cpp_int.hpp>
// namespace multiprecisioninteger = boost::multiprecision;
// using cint=multiprecisioninteger::cpp_int;
using namespace std;
using ll=long long;
using datas=pair<ll,ll>;
using ddatas=pair<long double,long double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
// using llset=tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>;
#define For(i,a,b) for(i=a;i<(ll)b;++i)
#define bFor(i,b,a) for(i=b,--i;i>=(ll)a;--i)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N,0)
#define brep1(i,N) bFor(i,N,1)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define uniq(v) vsort(v);(v).erase(unique(all(v)),(v).end())
#define endl "\n"
#define popcount __builtin_popcountll
#define eb emplace_back
#define print(x) cout<<x<<endl
#define printyes print("Yes")
#define printno print("No")
#define printYES print("YES")
#define printNO print("NO")
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
#define matoutput(v) do{for(auto outimat:v)output(outimat);}while(0)
constexpr ll mod=1000000007;
// constexpr ll mod=998244353;
constexpr ll inf=1LL<<60;
constexpr long double eps=1e-9;
const long double PI=acosl(-1);
template<class T,class E> ostream& operator<<(ostream& os,const pair<T,E>& p){return os<<"("<<p.first<<","<<p.second<<")";}
template<class T> ostream& operator<<(ostream& os,const vector<T>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T> ostream& operator<<(ostream& os,const set<T>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T,class E> ostream& operator<<(ostream& os,const map<T,E>& v){
  os<<"{";bool f=false;
  for(auto& x:v){if(f)os<<",";os<<x;f=true;}
  os<<"}";
  return os;
}
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;}
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;}
#ifdef DEBUG
void debugg(){cout<<endl;}
template<class T,class... Args>void debugg(const T& x,const Args&... args){cout<<" "<<x;debugg(args...);}
#define debug(...) cout<<__LINE__<<" ["<<#__VA_ARGS__<<"]:",debugg(__VA_ARGS__)
#else
#define debug(...) (void(0))
#endif

inline void startupcpp() noexcept{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(15);
}

ll modinv(ll a,ll m=mod) noexcept{
  ll b=m,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+m)%m;
}

ll moddevide(ll a,ll b) noexcept{return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(ll n,ll r) noexcept{
  if(n<r)return 0;
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1)modncrlistp[i]=modncrlistp[i-1]*i%mod;
    modncrlistm[n]=modinv(modncrlistp[n]);
    for(i=n;i>size;--i)modncrlistm[i-1]=modncrlistm[i]*i%mod;
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n,ll m=mod){
  if(n<0)return 0;
  ll res=1;
  while(n>0){
    if(n&1)res=res*a%m;
    a=a*a%m;
    n>>=1;
  }
  return res;
}

constexpr ll gcd(ll a,ll b) noexcept{return (!b)?abs(a):(a%b==0)?abs(b):gcd(b,a%b);}
constexpr ll lcm(ll a,ll b) noexcept{return a/gcd(a,b)*b;}

ll N;
string solve(string s,string t,int bit){
  // debug(s,t,bit);
  string ans;
  ans.reserve(N*3);
  int i=0,j=0;
  while(i<N*2||j<N*2){
    assert(i<=N*2&&j<=N*2);
    if(i==N*2){
      ans+=t[j++];
      continue;
    }
    if(j==N*2){
      ans+=s[i++];
      continue;
    }
    if(t[j]%2!=bit){
      ans+=t[j++];
      continue;
    }
    if(s[i]%2!=bit){
      ans+=s[i++];
      continue;
    }
    // if(s[i]!=t[j])debug(i,j,s[i],t[j]);
    assert(s[i]==t[j]);
    ans+=s[i];
    ++i;++j;
  }
  assert((int)ans.size()<=N*3);
  ans+=string(N*3-ans.size(),'0');
  return ans;
}
int main(){
  startupcpp();
int codeforces;cin>>codeforces;while(codeforces--){
  string s[3],ans;
  int i,j,k,cnt[3][2]={};
  cin>>N;
  rep(i,3)cin>>s[i];
  rep(j,3)rep(i,N*2)cnt[j][s[j][i]&1]++;
  rep(i,3)rep(j,i)rep(k,2){
    if(cnt[i][k]>=N&&cnt[j][k]>=N)ans=solve(s[i],s[j],k);
  }
  print(ans);
}
}