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. 
import java.util.*;
import java.util.Map.Entry;
import java.lang.*;
import java.math.*;
import java.text.*;
import java.io.*;
		public final class Solve {
		static List<List<Integer>> list = new ArrayList<>();
		static PrintWriter out = new PrintWriter(System.out); 
		static void flush() {
	 	out.flush();
	 	}
		static void run(long s, long e) {
		NumberFormat formatter = new DecimalFormat("#0.00000");
		System.out.print("Execution time is " + formatter.format((e - s) / 1000d) + " seconds");
		}
		
		static class FastReader {
	     static BufferedReader br;
	    StringTokenizer st;
	
	    public FastReader() {
	    br = new BufferedReader(new
	    InputStreamReader(System.in));
	    }
	    static boolean isPalindrome(String str1, String str2) {
	    String str3 = str1+str2;
	    int i = 0, j = str3.length()-1;
	    while(i < j) {
	    char a = str3.charAt(i), b = str3.charAt(j);
	    if(a != b) return false;
	    i++;j--;
	    }	
	    return true;
	    }
	    static boolean isPalindrome(String str) {
	    int i = 0, j = str.length()-1;
	    while(i < j) {
	    char a = str.charAt(i), b = str.charAt(j);
	    if(a != b) return false;
	    i++;j--;
	    }
	    return true;
	    }
	    String next() {
	    while (st == null || !st.hasMoreElements()) {
	    try{st = new StringTokenizer(br.readLine());}
	    catch (IOException  e) {e.printStackTrace();}
	    }
	    return st.nextToken();
	    }
	    int nextInt(){ return Integer.parseInt(next()); }
	    long nextLong() { return Long.parseLong(next());}
	    static int fact(int n) {
	    if(n == 1) return 1;
	    return n * fact(n-1);
	   	}
	    public int[] readIntArray(int n) {
	    int[] arr = new int[n];
		for(int i=0; i<n; ++i)
		arr[i]=nextInt();
		return arr;
		}
	    public long[] readLongArray(int n) {
	    	long[] arr = new long[n];
	    	for(int i=0; i<n; ++i)
	    		arr[i]=nextLong();
	    	return arr;
	    }
	    public int[][] readIntArray(int m, int n){
		int[][] arr = new int[m][n];
		for(int i = 0;i<m;i++) 
		for(int j = 0;j<n;j++) 
		arr[i][j] = nextInt();
		return arr;
	    }
	    public String[] readStringArray(int n) {
		String[] arr = new String[n];
		for(int i=0; i<n; ++i)
		arr[i]= nextLine();
		return arr;	
	    }
	    double nextDouble() {return Double.parseDouble(next());}
	    String nextLine() {
	    String str = "";
	    try{str = br.readLine();}
	    catch (IOException e) {e.printStackTrace();}
	    return str;}
	    }
		static void solve(int[] arr, int n) {
			
		}
		public static void main(String args[]) throws IOException {
	    FastReader sc = new FastReader();
	    long s1 = System.currentTimeMillis();
	    
	    int t = sc.nextInt();
	   loop: while(t-- > 0 ) {
	    String s = sc.nextLine();
	    String t1 = sc.nextLine();
	    String p = sc.nextLine();
	    
	    Set<Character> set = new HashSet<>();
	    
	    int[] arrs = new int[26];
	    int[] arrt = new int[26];
	    int[] arrp = new int[26];
	    for(int i = 0;i<t1.length();i++) {
	    	set.add(t1.charAt(i));
	    	arrt[t1.charAt(i) - 'a']++;
	    }
	    
	    for(int i = 0;i<s.length();i++) {
	    	arrs[s.charAt(i) - 'a']++;
	    }
	    for(int i = 0;i<p.length();i++) {
	    	arrp[p.charAt(i) - 'a']++;
	    }
	    int j = 0;
	    int fl = 0;
	    int i = 0;
	    for(i= 0;i < t1.length() && j < s.length();i++) {
	    	if(t1.charAt(i) != s.charAt(j)) {
	    		if(arrp[t1.charAt(i) - 'a'] > 0) {
	    			arrp[t1.charAt(i) - 'a']--;
	    		}
	    		else { 
	    			out.println("No");
	    			continue loop;
	    		}
	    	}
	    	else {
	    		j++;
	    	}
	    }
	    for(;i<t1.length();i++) {
	    	if(arrp[t1.charAt(i) - 'a'] > 0) {
    			arrp[t1.charAt(i) - 'a']--;
    		}
    		else { 
    			out.println("No");
    			continue loop;
    		}
	    }
	    	 if(j != s.length()) {
	 	    	out.println("No");
	 	    }
	    	 else
	    	out.println("Yes");
	    }
	    flush();
	    long e1 = System.currentTimeMillis();
//	    run(s1,e1);
	    }
	}
		
		
		