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:
/*
    Aman Agarwal
    algo.java
*/

import java.util.*;
import java.io.*;

public class C1194
{
    static class FastReader
    {
            BufferedReader br;
            StringTokenizer st;

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

            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());
            }

            double nextDouble()
            {
                return Double.parseDouble(next());
            }

            String nextLine()
            {
                String str = "";
                try
                {
                    str = br.readLine();
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
                return str;
            }
    }


    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int ni()throws IOException{return sc.nextInt();}
    static long nl()throws IOException{return sc.nextLong();}
    static int[][] nai2(int n,int m)throws IOException{int a[][] = new int[n][m];for(int i=0;i<n;i++)for(int j=0;j<m;j++)a[i][j] = ni();return a;}
    static int[] nai(int N,int start)throws IOException{int[]A=new int[N+start];for(int i=start;i!=(N+start);i++){A[i]=ni();}return A;}
    static Integer[] naI(int N,int start)throws IOException{Integer[]A=new Integer[N+start];for(int i=start;i!=(N+start);i++){A[i]=ni();}return A;}
    static long[] nal(int N)throws IOException{long[]A=new long[N];for(int i=0;i!=N;i++){A[i]=nl();}return A;}
    static char[] nac()throws IOException{char[]A=sc.next().toCharArray();return A;}
    static void print(int arr[]){for(int i=0;i<arr.length;i++)out.print(arr[i]+" ");out.println();}
    static void print(long arr[]){for(int i=0;i<arr.length;i++)out.print(arr[i]+" ");out.println();}
    static long gcd(long a, long b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b)throws IOException{return (b==0)?a:gcd(b,a%b);}
    static int bit(long n)throws IOException{return (n==0)?0:(1+bit(n&(n-1)));} // return the number of set bits.
    static boolean isPrime(int number){if(number==1) return false;if (number == 2 || number == 3){return true;}if (number % 2 == 0) {return false;}int sqrt = (int) Math.sqrt(number) + 1;for(int i = 3; i < sqrt; i += 2){if (number % i == 0){return false;}}return true;}
    static boolean isPrime(long number){if(number==1) return false;if (number == 2 || number == 3){return true;}if (number % 2 == 0) {return false;}long sqrt = (long) Math.sqrt(number) + 1;for(int i = 3; i < sqrt; i += 2){if (number % i == 0){return false;}}return true;}
    static int power(int n,int p){if(p==0)return 1;int a = power(n,p/2);a = a*a;int b = p & 1;if(b!=0){a =  n*a;}return a;}
    static long power(long n,long p){if(p==0)return 1;long a = power(n,p/2);a = a*a;long b = p & 1;if(b!=0){a =  n*a;}return a;}
    static void reverse(int[] a) {int b;for (int i = 0, j = a.length - 1; i < j; i++, j--) {b = a[i];a[i] = a[j];a[j] = b;}}
    static void reverse(long[] a) {long b;for (int i = 0, j = a.length - 1; i < j; i++, j--) {b = a[i];a[i] = a[j];a[j] = b;}}
    static void swap(int a[],int i,int j){a[i] = a[i] ^ a[j];a[j] = a[j] ^ a[i];a[i] = a[i] ^ a[j];}
    static void swap(long a[],int i,int j){a[i] = a[i] ^ a[j];a[j] = a[j] ^ a[i];a[i] = a[i] ^ a[j];}
    static int count(int n){int c=0;while(n>0){c++;n = n/10;}return c;}
    static int[] prefix_sum(int a[],int n){int s[] = new int[n];s[0] = a[0];for(int i=1;i<n;i++){s[i] = a[i]+s[i-1];}return s;}
    static long[] prefix_sum_int(int a[],int n){long s[] = new long[n];s[0] = (long)a[0];for(int i=1;i<n;i++){s[i] = ((long)a[i])+s[i-1];}return s;}
    static long[] prefix_sum_Integer(Integer a[],int n){long s[] = new long[n];s[0] = (long)a[0];for(int i=1;i<n;i++){s[i] = ((long)a[i])+s[i-1];}return s;}
    static long[] prefix_sum_long(long a[],int n){long s[] = new long[n];s[0] = a[0];for(int i=1;i<n;i++){s[i] = a[i]+s[i-1];}return s;}
    static boolean isPerfectSquare(double x){double sr = Math.sqrt(x);return ((sr - Math.floor(sr)) == 0);}
    static ArrayList<Integer> sieve(int n) {int k=0; boolean prime[] = new boolean[n+1];ArrayList<Integer> p_arr = new ArrayList<>();for(int i=0;i<n;i++) prime[i] = true;for(int p = 2; p*p <=n; p++){ k=p;if(prime[p] == true) { p_arr.add(p);for(int i = p*2; i <= n; i += p) prime[i] = false; } }for(int i = k+1;i<=n;i++){if(prime[i]==true)p_arr.add(i);}return p_arr;} 
    static boolean[] seive_check(int n) {boolean prime[] = new boolean[n+1];for(int i=0;i<n;i++) prime[i] = true;for(int p = 2; p*p <=n; p++){ if(prime[p] == true) { for(int i = p*2; i <= n; i += p) prime[i] = false; } }prime[1]=false;return prime;} 
    static int get_bits(int n){int p=0;while(n>0){p++;n = n>>1;}return p;}
    static int get_bits(long n){int p=0;while(n>0){p++;n = n>>1;}return p;}
    static int get_2_power(int n){if((n & (n-1))==0)return get_bits(n)-1;return -1;}
    static int get_2_power(long n){if((n & (n-1))==0)return get_bits(n)-1;return -1;}
    //gives the divisors of first n natural numbers in n*sqrt(n) time..
    static ArrayList<ArrayList<Integer>> divisors_n(int n){int i,j;ArrayList<ArrayList<Integer>> ans = new ArrayList<>();ArrayList<Integer> arr = new ArrayList<>();arr.add(1);ans.add(arr);for (i = 2; i <= n; i++){arr = new ArrayList<>();for (j = 1; j * j <= i; j++){if (i % j == 0){arr.add(j);if (i / j != j)arr.add(i/j);}}ans.add(arr);} return ans;} 
    //gives divisors of a particular number in sqrt(n) time..
    static ArrayList<Integer> divisors_1(int n){ArrayList<Integer> ans = new ArrayList<>();for (int i=1; i<=Math.sqrt(n); i++){if (n%i==0){if (n/i == i)ans.add(i);else{ans.add(i);ans.add(n/i);}}}return ans;} 
    static void close(){out.flush();}

    /*-------------------------Main Code Starts(algo.java)----------------------------------*/    

    public static void solve() throws IOException 
    {
        char s[] = sc.next().toCharArray();
        char t[] = sc.next().toCharArray();
        char p[] = sc.next().toCharArray();
        Map<Character,Integer> mp = new HashMap<>();
        Map<Character,Integer> mp1 = new HashMap<>();
        for(int i=0;i<p.length;i++)
        {
            if(mp.containsKey(p[i]))
            {
                int n = mp.get(p[i]);
                mp.put(p[i],n+1);
            }
            else
                mp.put(p[i],1);
        }

        int i=0,j=0,k=0,flg=0;
        char ans[] = new char[s.length+p.length+1];
        for(i=0;i<t.length;i++)
        {
            if(j<s.length && s[j]==t[i])
            {
                j++;
            }
            else
            {
                char x = t[i];
                if(!mp.containsKey(x) || mp.get(x)<=0)
                {
                    flg=1;
                    break;
                }
                else
                {
                    int y = mp.get(x);
                    mp.put(x,y-1);
                }
            }
        }
        if(flg==1 || j!=s.length)
            out.println("NO");
        else
            out.println("YES");

    }

    public static void main(String[] args) throws IOException 
    {
        int test = 1;
        test = sc.nextInt();
        while(test-- > 0)
        {
            solve();
        }
        close();
    }
}