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.BigInteger;
import java.util.*;

import static java.lang.Math.PI;
import static java.lang.System.in;
import static java.lang.System.out;
import static java.lang.System.err;

public class B { 
    public static void main(String[] args) throws Exception {   
        Foster sc = new Foster();
        PrintWriter p = new PrintWriter(out);
        int q = sc.nextInt();
        while(q--!=0){
            char S[] = sc.next().toCharArray();
            char T[] = sc.next().toCharArray();
            char P[] = sc.next().toCharArray();
            int freqS[] = new int[26];
            int freqT[] = new int[26];
            for(int i = 0; i < S.length; i++){
                int idx = (int)(S[i]-'a');
                freqS[idx]++;
            }
            for(int i = 0; i < P.length; i++){
                int idx = (int)(P[i]-'a');
                freqS[idx]++;
            }
            for(int i = 0; i < T.length; i++){
                int idx = (int)(T[i]-'a');
                freqT[idx]++;
            }
            boolean poss = true;
            for(int i = 0; i < 26; i++){
                if(freqS[i] < freqT[i]) poss = false;
            }
            boolean isSub = false;
            int idxS = 0, idxT = 0;
            while(idxS < S.length && idxT < T.length){
                if(S[idxS]==T[idxT]){
                    idxS++;
                }
                idxT++;
            }
            if(idxS==S.length)  isSub = true;
            p.println(poss && isSub ? "YES" : "NO");
        }
        p.close();
    }

    static int[] sort(int a[]){
        ArrayList<Integer> arr = new ArrayList<>();
        for(int i : a){
            arr.add(i);
        }
        Collections.sort(arr);
        for(int i = 0; i < arr.size(); i++){
            a[i] = arr.get(i);
        }
        return a;
    }
    
/* 
    3
    5 5 4 3
*/    
/*
1. Check overflow in pow function or in general
2. Check indices of read array function
3. Think of an easier solution because the problems you solve are always easy
4. Check iterator of loop
5. If still doesn't work, then jump from the 729th floor 'coz "beta tumse na ho paayega"

    Move to top!!
*/
    static class Foster {
        BufferedReader br = new BufferedReader(new InputStreamReader(in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                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());
        }
        int[] intArray(int n) {                   // Check indices
            int arr[] = new int[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }
        long[] longArray(int n) {                 // Check indices
            long arr[] = new long[n];
            for(int i = 0; i < n; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }
        int[] getBits(int n) {                   //in Reverse Order
            int a[] = new int[31];
            for(int i = 0; i < 31; i++) {
                if(((1<<i) & n) != 0)
                    a[i] = 1;
            }
            return a;
        }
        static String reverse(String s){
            String temp = "";
            for(int i = s.length()-1; i >= 0; i--){
                temp += s.charAt(i);
            }
            return temp;
        }
        static long pow(long... a) {
            long mod = Long.MAX_VALUE;
            if(a.length == 3)   mod = a[2];
            long res = 1;
            while(a[1] > 0) {
                if((a[1] & 1) == 1)
                    res = (res * a[0]) % mod;
                a[1] /= 2;
                a[0] = (a[0] * a[0]) % mod;
            }
            return res;
        }
        static void print(Object... o) {
            for(Object next : o) {
                System.err.print(next + " ");
            }
        }
        static void println(Object... o) {
            for(Object next : o) {
                System.err.print(next + " ");
            }
            System.err.println();
        }
        static void watch(Object...a) throws Exception {
            int i = 1;
            for (Object o: a) {
                boolean found = false;
                if (o.getClass().isArray()) {
                    String type = o.getClass().getName().toString();
                    switch (type) {
                        case "[I": {
                            int[] test = (int[]) o;
                            println(i + " : " + Arrays.toString(test));
                            break;
                        }
                        case "[[I": {
                            int[][] obj = (int[][]) o;
                            println(i + " : " + Arrays.deepToString(obj));
                            break;
                        }
                        case "[J": {
                            long[] obj = (long[]) o;
                            println(i + " : " + Arrays.toString(obj));
                            break;
                        }
                        case "[[J": {
                            long[][] obj = (long[][]) o;
                            println(i + " : " + Arrays.deepToString(obj));
                            break;
                        }
                        case "[D": {
                            double[] obj = (double[]) o;
                            println(i + " : " + Arrays.toString(obj));
                            break;
                        }
                        case "[[D": {
                            double[][] obj = (double[][]) o;
                            println(i + " : " + Arrays.deepToString(obj));
                            break;
                        }
                        case "[Ljava.lang.String": {
                            String[] obj = (String[]) o;
                            println(i + " : " + Arrays.toString(obj));
                            break;
                        }
                        case "[[Ljava.lang.String": {
                            String[][] obj = (String[][]) o;
                            println(i + " : " + Arrays.deepToString(obj));
                            break;
                        }
                        case "[C": {
                            char[] obj = (char[]) o;
                            println(i + " : " + Arrays.toString(obj));
                            break;
                        }
                        case "[[C": {
                            char[][] obj = (char[][]) o;
                            println(i + " : " + Arrays.deepToString(obj));
                            break;
                        }
                        default: {
                            println(i + " : type not identified");
                            break;
                        }
                    }
                    found = true;
                }
                if (o.getClass() == ArrayList.class) {
                    println(i + " : LIST = " + o);
                    found = true;
                }
                if (o.getClass() == TreeSet.class) {
                    println(i + " : SET = " + o);
                    found = true;
                }
                if (o.getClass() == TreeMap.class) {
                    println(i + " : MAP = " + o);
                    found = true;
                }
                if (o.getClass() == HashMap.class) {
                    println(i + " : MAP = " + o);
                    found = true;
                }
                if (o.getClass() == LinkedList.class) {
                    println(i + " : LIST = " + o);
                    found = true;
                }
                if (o.getClass() == PriorityQueue.class) {
                    println(i + " : PQ = " + o);
                    found = true;
                }
                if (!found) {
                    println(i + " = " + o);
                }
                i++;
            }
        }
    }
}