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.util.*;
import java.math.*;
import java.lang.*;
 
import static java.lang.Math.*;
 
public class Solution implements Runnable 
{
	static class InputReader 
	{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
 
		public InputReader(InputStream stream) 
		{
			this.stream = stream;
		}
		
		public int read()
		{
			if (numChars==-1) 
				throw new InputMismatchException();
            
			if (curChar >= numChars) 
			{
				curChar = 0;
				try
				{
					numChars = stream.read(buf);
				}
				catch (IOException e)
				{
					throw new InputMismatchException();
				}
                
				if(numChars <= 0)               
					return -1;
			}
			return buf[curChar++];
		}
     
		public String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}	
			return str;
		}
		public int nextInt() 
		{
			int c = read();
            
			while(isSpaceChar(c)) 
				c = read();
		
			int sgn = 1;
        
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
            
			int res = 0;
			do
			{
				if(c<'0'||c>'9') 
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}
			while (!isSpaceChar(c)); 
        
			return res * sgn;
		}
        
		public long nextLong() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-')
			{
				sgn = -1;
				c = read();
			}
			long res = 0;
			
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}	
			while (!isSpaceChar(c));
				return res * sgn;
		}
		
		public double nextDouble() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-')
			{
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') 
			{
				if (c == 'e' || c == 'E')
					return res * Math.pow(10, nextInt());
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') 
			{
				c = read();
				double m = 1;
				while (!isSpaceChar(c))
				{
					if (c == 'e' || c == 'E')
						return res * Math.pow(10, nextInt());
					if (c < '0' || c > '9')
						throw new InputMismatchException();
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
    
		public String readString() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} 
			while (!isSpaceChar(c));
            
			return res.toString();
		}
     
		public boolean isSpaceChar(int c) 
		{
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
     
		public String next()
		{
			return readString();
		}
        
		public interface SpaceCharFilter
		{
			public boolean isSpaceChar(int ch);
		}
		
		public int[] readint(int st,int n){
		    int[] arr=new int[n+st];
		    for(int i=st;i<n+st;i++) arr[i]=nextInt();
		    return arr;
		}
		
		public long[] readlong(int st,int n){
		    long[] arr=new long[n+st];
		    for(int i=st;i<n+st;i++) arr[i]=nextLong();
		    return arr;
		}
	}
	public static void main(String args[]) throws Exception 
	{
		new Thread(null, new Solution(),"Main",1<<27).start();
	}	
	
	public static long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b % a, a); 
	} 
  
	public static long findGCD(long arr[], int n) 
	{ 
		long result = arr[0]; 
		for (int i = 1; i < n; i++) 
			result = gcd(arr[i], result); 
		return result; 
	}	 
	
	static void sortbycolomn(long arr[][], int col)
    { 
         
        Arrays.sort(arr, new Comparator<long[]>() { 
            
          @Override              
           
          public int compare(final long[] entry1,  
                             final long[] entry2) { 
  
            
            if (entry1[col] > entry2[col]) 
                return 1; 
            else
                return -1; 
          } 
        });   
    } 
	
	public void run()
	{
		InputReader in = new InputReader(System.in);
		PrintWriter w = new PrintWriter(System.out);
		
		int k=in.nextInt();
		while(k--!=0){
		    char[] s=in.next().toCharArray();
		    char[] t=in.next().toCharArray();
		    char[] p=in.next().toCharArray();
		    int ss=s.length,tt=t.length,pp=p.length;
		    int[] has=new int[26];
		    for(int i=0;i<pp;i++) has[p[i]-97]++;
		    
		    //w.println(Arrays.toString(has));
		    
		    int i=0,j=0,flag=0;
		    for(i=0,j=0;i<tt && j<ss;i++){
		        if(t[i]==s[j]) j++;   
		        else if(has[t[i]-97]>0){
		            has[t[i]-97]--;
		        }else{
		            flag=1;
		            break;
		        }
		    }
		    if(flag==1 || j<ss){
		        w.println("NO");
		        continue;
		    }
		    while(i<tt){
		        if(has[t[i]-97]>0)  has[t[i]-97]--;
		        else{
		            flag=1;
		            break;
		        }
		        i++;
		    }
		    //w.println(i);
		    if(flag==1){
		        w.println("NO");
		        continue;
		    }
		    w.println("YES");
		}
		
		w.flush();
		w.close();
	}
}