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.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.LinkedBlockingQueue;
import java.util.zip.Inflater;

public class Main {
    private static AWriter writer = new AWriter(System.out);
    private static AReader reader = new AReader(System.in);
    private static final int INF=0X3F3F3F3F;
    private static int n,m;
    private static char[] s,t,p;
    private static int dp[][];
    public static void main(String[] args) throws IOException {
        int T=reader.nextInt();
        while(T--!=0) {
            s=reader.next().toCharArray();
            t=reader.next().toCharArray();
            p=reader.next().toCharArray();
            int[] cnt1=new int[29],cnt2=new int[29],cnt3=new int[29];
            boolean flag=true;
            int i,j;
            for(i=0,j=0;i<t.length&&j<s.length;i++)
                if(s[j]==t[i])
                    j++;
            if(j!=s.length)
                flag=false;
            if(!flag) {
                writer.println("NO");
                continue;
            }
            for(i=0;i<s.length;i++)
                cnt1[s[i]-'a']++;
            for(i=0;i<t.length;i++)
                cnt2[t[i]-'a']++;
            for(i=0;i<p.length;i++)
                cnt3[p[i]-'a']++;
            for(i=0;i<29;i++){
                if(cnt1[i]+cnt3[i]<cnt2[i]) {
                    flag = false;
                    break;
                }
            }
            if(flag)
                writer.println("YES");
            else
                writer.println("NO");
        }
        writer.close();
        reader.close();
    }
}

class AReader implements Closeable {
    private BufferedReader reader;
    private StringTokenizer tokenizer;
    public AReader(InputStream inputStream) {
        reader = new BufferedReader(new InputStreamReader(inputStream));
        tokenizer = new StringTokenizer("");
    }
    private String innerNextLine() {
        try {
            return reader.readLine();
        } catch (IOException ex) {
            return null;
        }
    }
    public boolean hasNext() {
        while (!tokenizer.hasMoreTokens()) {
            String nextLine = innerNextLine();
            if (nextLine == null) {
                return false;
            }
            tokenizer = new StringTokenizer(nextLine);
        }
        return true;
    }
    public String nextLine() {
        tokenizer = new StringTokenizer("");
        return innerNextLine();
    }
    public String next() {
        hasNext();
        return tokenizer.nextToken();
    }
    public int nextInt() {
        return Integer.valueOf(next());
    }
    public long nextLong() {
        return Long.valueOf(next());
    }
    public double nextDouble() {
        return Double.valueOf(next());
    }
    public BigDecimal nextBigDecimal(){ return new BigDecimal(next()); }
    public BigInteger nextBigInteger() {
        return new BigInteger(next());
    }
    @Override
    public void close() throws IOException {
        reader.close();
    }
}

// Fast writer for ACM By Azure99
class AWriter implements Closeable {
    private BufferedWriter writer;

    public AWriter(OutputStream outputStream) {
        writer = new BufferedWriter(new OutputStreamWriter(outputStream));
    }

    public void print(Object object) throws IOException {
        writer.write(object.toString());
    }

    public void println(Object object) throws IOException {
        writer.write(object.toString());
        writer.write("\n");
    }

    @Override
    public void close() throws IOException {
        writer.close();
    }
}