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.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        Scanner in = new Main().new Scanner(inputStream);
        Main solver = new Main();
        solver.runn(in);
    }

    public void runn(Scanner in) {
        int q = in.ni();
        while (q-- > 0) {
            String s = in.rl(), t = in.rl(), p = in.rl();
            int[] charCount = new int[26];
            for (char c : p.toCharArray())
                charCount[c - 'a']++;
            int i = 0, j = 0;
            while (i < s.length() || j < t.length()) {
                if (i >= s.length()) {
                    if (charCount[t.charAt(j) - 'a'] > 0) {
                        charCount[t.charAt(j) - 'a']--;
                        j++;
                    } else
                        break;
                } else if (j >= t.length())
                    break;
                else if (s.charAt(i) == t.charAt(j)) {
                    i++;
                    j++;
                } else if (charCount[t.charAt(j) - 'a'] > 0) {
                    charCount[t.charAt(j) - 'a']--;
                    j++;
                } else
                    break;
            }
            boolean good = (i == s.length() && j == t.length());
            if (!good) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
            }
        }
    }

    class Scanner {
        BufferedReader br;
        StringTokenizer in;

        public Scanner(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            return hasMoreTokens() ? in.nextToken() : null;
        }

        public String rl() {
            try {
                return br.readLine();
            } catch (Exception e) {
                e.printStackTrace();
                return null;
            }
        }

        boolean hasMoreTokens() {
            while (in == null || !in.hasMoreTokens()) {
                String s = rl();
                if (s == null)
                    return false;
                in = new StringTokenizer(s);
            }
            return true;
        }

        public String nextString() {
            return next();
        }

        public int ni() {
            return Integer.parseInt(nextString());
        }

        public double nd() {
            return Double.parseDouble(nextString());
        }

        public long nl() {
            return Long.parseLong(nextString());
        }

        public int[] nia(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nla(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }
    }

    class Pair implements Comparable<Pair> {
        long _first, _second;

        public Pair(int f, int s) {
            _first = f;
            _second = s;
        }

        public Pair(Long f, Long s) {
            _first = f;
            _second = s;
        }

        public int compareTo(Pair o) {
            if (this.first() != o.first())
                return Long.compare(this.first(), o.first());
            else
                return Long.compare(this.second(), o.second());
        }

        long first() {
            return _first;
        }

        long second() {
            return _second;
        }

        int firsti() {
            return (int) _first;
        }

        int secondi() {
            return (int) _second;
        }
    }

    class Triple implements Comparable<Triple> {
        Integer _first, _second, _third;

        public Triple(Integer f, Integer s, Integer t) {
            _first = f;
            _second = s;
            _third = t;
        }

        public int compareTo(Triple o) {
            if (!this.first().equals(o.first()))
                return this.first() - o.first();
            else if (!this.second().equals(o.second()))
                return this.second() - o.second();
            else
                return this.third() - o.third();
        }

        Integer first() {
            return _first;
        }

        Integer second() {
            return _second;
        }

        Integer third() {
            return _third;
        }

        public String toString() {
            return first() + " " + second() + " " + third();
        }
    }
}