Problem: You are given three strings s, t and p consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.

During each operation you choose any character from p, erase it from p and insert it into string s (you may insert this character anywhere you want: in the beginning of s, in the end or between any two consecutive characters). 

For example, if p is aba, and s is de, then the following outcomes are possible (the character we erase from p and insert into s is highlighted):

  * aba → ba, de → ade; 
  * aba → ba, de → dae; 
  * aba → ba, de → dea; 
  * aba → aa, de → bde; 
  * aba → aa, de → dbe; 
  * aba → aa, de → deb; 
  * aba → ab, de → ade; 
  * aba → ab, de → dae; 
  * aba → ab, de → dea; 



Your goal is to perform several (maybe zero) operations so that s becomes equal to t. Please determine whether it is possible.

Note that you have to answer q independent queries.

Input

The first line contains one integer q (1 ≤ q ≤ 100) — the number of queries. Each query is represented by three consecutive lines.

The first line of each query contains the string s (1 ≤ |s| ≤ 100) consisting of lowercase Latin letters.

The second line of each query contains the string t (1 ≤ |t| ≤ 100) consisting of lowercase Latin letters.

The third line of each query contains the string p (1 ≤ |p| ≤ 100) consisting of lowercase Latin letters.

Output

For each query print YES if it is possible to make s equal to t, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


4
ab
acxb
cax
a
aaaa
aaabbcc
a
aaaa
aabbcc
ab
baaa
aaaaa


Output


YES
YES
NO
NO

Note

In the first test case there is the following sequence of operation: 

  1. s =  ab, t =  acxb, p =  cax; 
  2. s =  acb, t =  acxb, p =  ax; 
  3. s =  acxb, t =  acxb, p =  a. 



In the second test case there is the following sequence of operation: 

  1. s =  a, t =  aaaa, p =  aaabbcc; 
  2. s =  aa, t =  aaaa, p =  aabbcc; 
  3. s =  aaa, t =  aaaa, p =  abbcc; 
  4. s =  aaaa, t =  aaaa, p =  bbcc. 

Solution (in JAVA):

Solution:

// template halnix16


import java.io.*;
import java.util.*;
public class Main {


    static BufferedReader br;

    public static void main(String[] args) throws Exception {
        // write your code here

        input input = new input();
        int tc = input.in();

        while (tc-- > 0) {
            String s, t, p;
            s = input.sin();
            t = input.sin();
            p = input.sin();

            int tl = t.length();
            int sl = s.length();
            if (tl < sl) {
                System.out.println("NO");
                continue;
            }
            if (tl == sl) {
                if (s.equals(t)) {
                    System.out.println("YES");
                    continue;
                }
                System.out.println("NO");
                continue;
            }
            HashMap<Character, Integer> pf = new HashMap<>();
            for (char c : p.toCharArray()) {

                int f = pf.containsKey(c) ? pf.get(c) : 0;
                pf.put(c, f + 1);
            }

            int si = 0;
            String ans = "YES";
            for (char c : t.toCharArray()) {

                if(si<sl) {
                    char sc = s.charAt(si);
                    if(sc==c){
                        si+=1;
                        continue;
                    } else {
                        if(pf.containsKey(c)){
                            int f = pf.get(c);
                            pf.put(c,f-1);
                            if(f==1){
                                pf.remove(c);
                            }
                            continue;
                        } else{
                            ans="NO";
                            break;
                        }
                    }

                }
                if(pf.containsKey(c)){
                    int f = pf.get(c);
                    pf.put(c,f-1);
                    if(f==1){
                        pf.remove(c);
                    }
                    continue;
                } else{
                    ans="NO";
                    break;
                }
            }
            if(si>=sl){
                System.out.println(ans);
            }else{
                System.out.println("NO");
            }
        }
    }
}





class input{

    BufferedReader br ;
    public input(){
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    int in() throws Exception{
        return Integer.parseInt(br.readLine());
    }

    long Lin() throws Exception{
        return Long.parseLong(br.readLine());


    }
    long Lin(String num) throws Exception{
        return Long.parseLong(num);


    }


    String sin() throws Exception{

        return br.readLine();
    }

    String [] sarrin() throws Exception{

        return br.readLine().split(" ");

    }

    String [] regexSplit() throws Exception{
        return br.readLine().split(" ?(?<!\\G)((?<=[^\\p{Punct}])(?=\\p{Punct})|\\b) ?");
    }
    int in(String num) throws Exception{

        return Integer.parseInt(num);
    }

}