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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CFromSToT solver = new CFromSToT();
        int testCount = Integer.parseInt(in.next());
        for (int i = 1; i <= testCount; i++)
            solver.solve(i, in, out);
        out.close();
    }

    static class CFromSToT {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            char[] s = in.nextString().toCharArray();
            char[] t = in.nextString().toCharArray();
            char[] p = in.nextString().toCharArray();
            int lenLCS = DPUtils.lcs(s, t);
            if (lenLCS != s.length) {
                out.println("NO");
                return;
            }
            int[] count1 = StringUtils.countAll(new String(s) + new String(p));
            int[] count2 = StringUtils.countAll(new String(t));
            boolean flag = true;
            for (int i = 0; i < count1.length; ++i) {
                flag &= count1[i] >= count2[i];
            }
            out.println(flag ? "YES" : "NO");
        }

    }

    static class DPUtils {
        public static int lcs(char[] s, char[] t) {
            final int m = s.length;
            final int n = t.length;
            int[][] dp = new int[m + 1][n + 1];
            for (int i = 0; i <= m; i++) {
                for (int j = 0; j <= n; j++) {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 0;
                    } else if (s[i - 1] == t[j - 1]) {
                        dp[i][j] = dp[i - 1][j - 1] + 1;
                    } else {
                        dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                    }
                }
            }
            return dp[m][n];
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        private int pread() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

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

        public String nextString() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = pread();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }

    static class StringUtils {
        public static int[] countAll(String string) {
            int[] c = new int[128];
            for (char s : string.toCharArray()) {
                c[s]++;
            }
            return c;
        }

    }
}