#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
#define debug(x) cout<<#x<<": "<<x<<endl
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
using ll=long long;
using ull=unsigned long long;
void solve(){
    string a,b,c;
    cin>>a>>b>>c;
    if(a.size()!=b.size()){
        cout<<"NO"<<endl;
        return;
    }
    vector<int> f(26);
    for(int i=0;i<26;i++)
        f[i]=i;
    auto get=[&](int x){
        while(x!=f[x])
            x=f[x]=f[f[x]];
        return x;
    };
    auto merge=[&](int x,int y)->void{
        x=get(x),y=get(y);
        if(x==y)
            return;
        f[y]=x;
    };
    for(int i=0;i<a.size();i++){
        if(a[i]!=b[i]){
            int x=a[i]-'a',y=b[i]-'a';
            merge(x,y);
        }
    }
    if(c.size()!=a.size()){
        cout<<"YES"<<endl;
        return;
    }
    for(int i=0;i<a.size();i++){
        int cur=a[i]-'a';
        cur=get(cur);
        a[i]='a'+cur;
    }
    for(int i=0;i<c.size();i++){
        int cur=c[i]-'a';
        cur=get(cur);
        c[i]='a'+cur;
    }   
    if(a==c)
        cout<<"NO"<<endl;
    else
        cout<<"YES"<<endl; 
}
signed main(){
    ios::sync_with_stdio(0);
    cout.tie(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--){
        solve();
    }
}
/*
贡献法
正难则反
数小状压
关系连边
拆位
广义单调性
最长转最短
*/