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.util.NoSuchElementException;


public class Main_1194C {

	private static Scanner sc;
	private static Printer pr;

	private static void solve() {
		int q = sc.nextInt();

		out:
		for (int tcase = 0; tcase < q; tcase++) {
			char[] s = sc.next().toCharArray();
			char[] t = sc.next().toCharArray();
			char[] p = sc.next().toCharArray();
			
			int[] cntp = new int[26];
			for (char c : p) {
				cntp[c - 'a']++;
			}

			int j = 0;
			for (int i = 0, size = t.length; i < size; i++) {
				if (j < s.length && t[i] == s[j]) {
					j++;
				} else {
					if (cntp[t[i] - 'a'] == 0) {
						pr.println("NO");
						continue out;
					} else {
						cntp[t[i] - 'a']--;
					}
				}
			}

			if (j == s.length) {
				pr.println("YES");
			} else {
				pr.println("NO");
			}
		}
	}

	// ---------------------------------------------------
	public static void main(String[] args) {
		sc = new Scanner(System.in);
		pr = new Printer(System.out);
			
		solve();
			
		pr.close();
		sc.close();
	}

	static class Scanner {
		BufferedReader br;

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

		private boolean isPrintable(int ch) {
			return ch >= '!' && ch <= '~';
		}

		private boolean isCRLF(int ch) {
			return ch == '\n' || ch == '\r' || ch == -1;
		}

		private int nextPrintable() {
			try {
				int ch;
				while (!isPrintable(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}

				return ch;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		String next() {
			try {
				int ch = nextPrintable();
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (isPrintable(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		int nextInt() {
			try {
				// parseInt from Integer.parseInt()
				boolean negative = false;
				int res = 0;
				int limit = -Integer.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Integer.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				int multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		long nextLong() {
			try {
				// parseLong from Long.parseLong()
				boolean negative = false;
				long res = 0;
				long limit = -Long.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Long.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				long multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

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

		String nextLine() {
			try {
				int ch;
				while (isCRLF(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (!isCRLF(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}
		
		int[] nextIntArray(int n) {
			int[] ret = new int[n];
			for (int i = 0; i < n; i++) {
				ret[i] = sc.nextInt();
			}
			
			return ret;
		}

		long[] nextLongArray(int n) {
			long[] ret = new long[n];
			for (int i = 0; i < n; i++) {
				ret[i] = sc.nextLong();
			}
			
			return ret;
		}

		int[][] nextIntArrays(int n, int m) {
			int[][] ret = new int[m][n];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					ret[j][i] = sc.nextInt();
				}
			}
			
			return ret;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
//				throw new NoSuchElementException();
			}
		}
	}

	static class Printer extends PrintWriter {
		Printer(OutputStream out) {
			super(out);
		}
		
		void printInts(int... a) {
			StringBuilder sb = new StringBuilder(32);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}
		
		void printLongs(long... a) {
			StringBuilder sb = new StringBuilder(64);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}
		
		void printStrings(String... a) {
			StringBuilder sb = new StringBuilder(32);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}
	}
}