description: You are given two strings s and t, both consisting of lowercase English letters. You are going to type the string s character by character, from the first character to the last one.
  
  When typing a character, instead of pressing the button corresponding to it, you can press the "Backspace" button. It deletes the last character you have typed among those that aren't deleted yet (or does nothing if there are no characters in the current string). For example, if s is "abcbd" and you press Backspace instead of typing the first and the fourth characters, you will get the string "bd" (the first press of Backspace deletes no character, and the second press deletes the character 'c'). Another example, if s is "abcaa" and you press Backspace instead of the last two letters, then the resulting text is "a".
  
  Your task is to determine whether you can obtain the string t, if you type the string s and press "Backspace" instead of typing several (maybe zero) characters of s.
  
  Input
  
  The first line contains a single integer q (1 ≤ q ≤ 10^5) — the number of test cases.
  
  The first line of each test case contains the string s (1 ≤ |s| ≤ 10^5). Each character of s is a lowercase English letter.
  
  The second line of each test case contains the string t (1 ≤ |t| ≤ 10^5). Each character of t is a lowercase English letter.
  
  It is guaranteed that the total number of characters in the strings over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case, print "YES" if you can obtain the string t by typing the string s and replacing some characters with presses of "Backspace" button, or "NO" if you cannot.
  
  You may print each letter in any case (YES, yes, Yes will all be recognized as positive answer, NO, no and nO will all be recognized as negative answer).
  
  Example
  
  Input
  
  
  4
  ababa
  ba
  ababa
  bb
  aaa
  aaaa
  aababa
  ababa
  
  
  Output
  
  
  YES
  NO
  NO
  YES
  
  Note
  
  Consider the example test from the statement.
  
  In order to obtain "ba" from "ababa", you may press Backspace instead of typing the first and the fourth characters.
  
  There's no way to obtain "bb" while typing "ababa".
  
  There's no way to obtain "aaaa" while typing "aaa".
  
  In order to obtain "ababa" while typing "aababa", you have to press Backspace instead of typing the first character, then type all the remaining characters.
solution: import java.util.*;
  import java.io.*;
  
  public class B1557 {
  	static boolean multipleTC = true;
  	FastReader in;
  	PrintWriter out;
  
  	static int mod = 1000000007;
  
  	public static void main(String[] args) throws Exception {
  		new B1557().run();
  	}
  
  	void run() throws Exception {
  		in = new FastReader();
  		out = new PrintWriter(System.out);
  		int T = (multipleTC) ? ni() : 1;
  		pre();
  		for (int t = 1; t <= T; t++)
  			solve(t);
  		out.flush();
  		out.close();
  	}
  
  	void pre() throws Exception {
  	}
  
  	void solve(int TC) throws Exception {
  		String s = nln();
  		String t = nln();
  		int ptr1 = s.length() - 1;
  		int ptr2 = t.length() - 1;
  		while (ptr1 >= 0 && ptr2 >= 0) {
  			if (s.charAt(ptr1) == t.charAt(ptr2)) {
  				ptr1--;
  				ptr2--;
  			} else {
  				ptr1 -= 2;
  			}
  		}
  		if (ptr2 == -1) {
  			pn("YES");
  		} else {
  			pn("NO");
  		}
  
  	}
  
  	static int bitCount(int x) {
  		return x == 0 ? 0 : (1 + bitCount(x & (x - 1)));
  	}
  
  	static void dbg(Object... o) {
  		System.err.println(Arrays.deepToString(o));
  	}
  
  	public static boolean isPrime(long n) {
  		if (n < 2)
  			return false;
  		if (n == 2 || n == 3)
  			return true;
  		if (n % 2 == 0 || n % 3 == 0)
  			return false;
  		long sqrtN = (long) Math.sqrt(n) + 1;
  		for (long i = 6L; i <= sqrtN; i += 6) {
  			if (n % (i - 1) == 0 || n % (i + 1) == 0)
  				return false;
  		}
  		return true;
  	}
  
  	public static long gcd(long a, long b) {
  		if (a > b)
  			a = (a + b) - (b = a);
  		if (a == 0L)
  			return b;
  		return gcd(b % a, a);
  	}
  
  	public long pow(long base, long exp, long MOD) {
  		base %= MOD;
  		long res = 1;
  		while (exp > 0) {
  			if ((exp & 1) == 1)
  				res = res * base % MOD;
  			base = base * base % MOD;
  			exp >>= 1;
  		}
  		return res;
  	}
  
  	public static long totient(long n) {
  		long result = n;
  		for (int p = 2; p * p <= n; ++p)
  			if (n % p == 0) {
  				while (n % p == 0)
  					n /= p;
  				result -= result / p;
  			}
  		if (n > 1)
  			result -= result / n;
  		return result;
  	}
  
  	int bit(long n) {
  		return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
  	}
  
  	public long max(long... arr) {
  		long max = arr[0];
  		for (long itr : arr)
  			max = Math.max(max, itr);
  		return max;
  	}
  
  	public int max(int... arr) {
  		int max = arr[0];
  		for (int itr : arr)
  			max = Math.max(max, itr);
  		return max;
  	}
  
  	public long min(long... arr) {
  		long min = arr[0];
  		for (long itr : arr)
  			min = Math.min(min, itr);
  		return min;
  	}
  
  	public int min(int... arr) {
  		int min = arr[0];
  		for (int itr : arr)
  			min = Math.min(min, itr);
  		return min;
  	}
  
  	public long sum(long... arr) {
  		long sum = 0;
  		for (long itr : arr)
  			sum += itr;
  		return sum;
  	}
  
  	public long sum(int... arr) {
  		long sum = 0;
  		for (int itr : arr)
  			sum += itr;
  		return sum;
  	}
  
  	void p(Object o) {
  		out.print(o);
  	}
  
  	void pn(Object o) {
  		out.println(o);
  	}
  
  	void pni(Object o) {
  		out.println(o);
  		out.flush();
  	}
  
  	String n() throws Exception {
  		return in.next();
  	}
  
  	String nln() throws Exception {
  		return in.nextLine();
  	}
  
  	int ni() throws Exception {
  		return Integer.parseInt(in.next());
  	}
  
  	long nl() throws Exception {
  		return Long.parseLong(in.next());
  	}
  
  	double nd() throws Exception {
  		return Double.parseDouble(in.next());
  	}
  
  	class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		public FastReader(String s) throws Exception {
  			br = new BufferedReader(new FileReader(s));
  		}
  
  		String next() throws Exception {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					throw new Exception(e.toString());
  				}
  			}
  			return st.nextToken();
  		}
  
  		String nextLine() throws Exception {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				throw new Exception(e.toString());
  			}
  			return str;
  		}
  	}
  }