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.math.BigInteger;
import java.util.*;
import java.io.*;

public class solution {
    public static void merge(int arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    public static void sort(int arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    public static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        } else return gcd(b, a % b);
    }

    static class pair {
        int x;
        int y;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public static long lcm(long a, long b) {
        long v = a * b;
        long u = gcd(a, b);
        return v / u;
    }

    public static int binomialCoeff(int n, int k) {
        int C[] = new int[k + 1];

        // nC0 is 1
        C[0] = 1;

        for (int i = 1; i <= n; i++) {
            // Compute next row of pascal
            // triangle using the previous row
            for (int j = Math.min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        for (int ii = 0; ii < n; ii++) {
            StringBuilder s = new StringBuilder(br.readLine());
            StringBuilder t = new StringBuilder(br.readLine());
            StringBuilder p = new StringBuilder(br.readLine());
            if(s.length()<t.length()){
                for(int i = 0;i<t.length()-s.length();i++){
                    s.append('1');
                }
            }
            boolean flag = true;
            for (int i = 0; i <s.length(); i++) {
                if (t.length() <= i) {
                    if(s.charAt(i)!='1')
                    flag = false;
                    break;
                } else {
                    if (s.charAt(i) != t.charAt(i)) {
                        if (p.length() == 0) {
                            flag = true;
                        }
                        boolean f = false;
                        for (int j = 0; j < p.length(); j++) {
                            if (p.charAt(j) == t.charAt(i)) {
                                s.insert(i, p.charAt(j));
                                p.deleteCharAt(j);
                                f = true;
                                break;
                            }
                        }
                        if (f == false) {
                            flag = false;
                            break;
                        }
                    }
                }
            }
            int i = s.length()-1;
            while (s.charAt(i) == '1'){
                s.deleteCharAt(i);
                i--;
            }
            if (flag && (s.length() == t.length())) {
                System.out.println("YES");
            } else {
                System.out.println("NO");
            }
        }
    }

    static class sort implements Comparator<ArrayList<Integer>> {

        @Override
        public int compare(ArrayList<Integer> o1, ArrayList<Integer> o2) {
            int c = o2.get(1).compareTo(o1.get(1));
            if (c == 0) {
                c = o2.get(0).compareTo(o1.get(0));
            }
            return c;
        }

    }
}