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:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
 
/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader sc = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, sc, out);
        out.close();
    }
 
    static class Task {
        public void solve(int testNumber, InputReader sc, PrintWriter out) {
            int q=sc.nextInt();
            
            while(q-->0) {
            	String s=sc.next();
            	String t=sc.next();
            	String p=sc.next();
            	Mac jud=new Mac(t);
            	
            	if(!jud.check(s)) {
            		out.println("NO");
            		continue;
            	}
            	int[] cnt_p=new int[26];
            	int[] cnt_s=new int[26];
            	int[] cnt_t=new int[26];
            	for(int i=0;i<p.length();i++)
            		cnt_p[p.charAt(i)-'a']++;
            	for(int i=0;i<s.length();i++)
            		cnt_s[s.charAt(i)-'a']++;
                for(int i=0;i<t.length();i++)
                	cnt_t[t.charAt(i)-'a']++;
                for(int i=0;i<26;i++)
                	cnt_s[i]=cnt_t[i]-cnt_s[i];
                boolean flag=false;
                for(int i=0;i<26;i++) {
                	if(cnt_p[i]<cnt_s[i]) {
                		flag=true;
                		break;
                	}
                }
                if(flag) {
                	out.println("NO");
                	continue;
                }
                out.println("YES");
            }
        }
 
    }
    
    static class Mac{
        int[][] next;
        String temp;
         
        public Mac(String temp) {
            this.temp=temp;
            next=new int[temp.length()+1][26];
            insert();
        }
         
        public void insert() {
            int n=temp.length();
            for(int i=n;i>=1;i--) {
                for(int j=0;j<26;j++) {
                    next[i-1][j]=next[i][j];
                }
                if(i<n)
                    next[i-1][temp.charAt(i)-'a']=i;
            }
        }
         
        public boolean check(String res) {
            int index=0;
            int n=res.length();
            int k=0;
            if(res.charAt(index)==temp.charAt(0))
                index++;
             
            while(index<n) {
                if(next[k][res.charAt(index)-'a']!=0) {
                    k=next[k][res.charAt(index)-'a'];
                    index++;
                }
                else
                    return false;
            }
            return index>=n;
        }
    }
 
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
 
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
 
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(next());
        }
 
        public long nextLong() {
            return Long.parseLong(next());
        }
 
    }
}