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.
//package cp;
import java.util.*;


import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.max;
import java.io.*;
 
public class Codechef {
	static long[] mods = {1000000007, 998244353, 1000000009};
	static long mod = mods[0];
	public static MyScanner in;
    public static PrintWriter out;
    static int[][] dp;
    
    
	public static void main( String[] args ) throws IOException{
		
 		in = new MyScanner();
 		out = new PrintWriter(System.out);
 		StringBuilder ans1=new StringBuilder();
		int test=in.nextInt();
		while(test-->0) {
			//StringBuilder ans=new StringBuilder();
			char[] s=in.next().toCharArray(); 
			char[] t=in.next().toCharArray();
			
			boolean flag=true;
			int i=s.length-1;
			String ans="";
			
			for(int j=t.length-1;j>=0;j--){
			     
			     while(i>=0 && s[i]!=t[j]){
			          i-=2;
			     }
			     if(i<0){
			          flag=false;
			          break;
			     }
			     i-=1;
			     
			}
			
			if(flag)  ans="YES";
			else      ans="NO";
			
			ans1.append(ans).append(System.getProperty("line.separator"));
		}
		out.println(ans1);
 		out.close();
 		
 	}
	
	static long lcm_comp(long a, long b) {	return (a / gcd_comp(a, b)) * b;	}
 
	static long gcd_comp(long a, long b){
        if (a == 0)		return b;
        if (b == 0)		return a;
        long k;
        for (k = 0; ((a | b) & 1) == 0; ++k){
            a >>= 1;
            b >>= 1;
        }
        while ((a & 1) == 0)	a >>= 1;  
        do{
            while ((b & 1) == 0)	 b >>= 1;
            if (a > b){    
                long temp = a;
                a = b;
                b = temp;
            }
            b = (b - a);
        } while (b != 0);
        return a << k;
    }

	
	//comparator to solve according to b in descending order
	static class Pair implements Comparable<Pair>{
		private long a;
		private long b;
		public Pair(long a, long b) {
			this.a = a;
			this.b = b;
		}
		public int compareTo(Pair other) {
			return Long.compare(this.b, other.b );		//to sort in ascending order swap the arguments in Long.compare
			//return Long.compare(other.b ,this.b); 	//to sort in descending order swap the arguments in Long.compare
		}
	}
	
	static void printArray(long[] a) {
		StringBuilder stringBuilder=new StringBuilder();
		for(long i: a)	stringBuilder.append(i).append(" ");
		out.println(stringBuilder);
	}
	
	public static void sort(long[] array){
		ArrayList<Long> copy = new ArrayList<>();
		for (long i : array)
			copy.add(i);
		Collections.sort(copy);
		for(int i = 0;i<array.length;i++)
			array[i] = copy.get(i);
	}
	
	public static void sortDes(long[] array){
		ArrayList<Long> copy = new ArrayList<>();
		for (long i : array)
			copy.add(i);
		Collections.sort(copy);
		int j=array.length-1;
		for(int i = 0;i<array.length;i++)
			array[j--] = copy.get(i);
	}
	
	
	static String[] readArrayString(int n){
		String[] array = new String[n];
		for(int j =0 ;j<n;j++)
			array[j] = in.next();
		return array;
	}
	
	
	static int[] readArrayInt(int n){
    	int[] array = new int[n];
    	for(int j = 0;j<n;j++)
    		array[j] = in.nextInt();
    	return array;
    }
	
	static int[] readArrayInt1(int n){
		int[] array = new int[n+1];
		for(int j = 1;j<=n;j++){
			array[j] = in.nextInt();
		}
		return array;
	}
	
	static long[] readArrayLong(int n){
		long[] array = new long[n];
		for(int j =0 ;j<n;j++)
			array[j] = in.nextLong();
		return array;
	}
	
	static double[] readArrayDouble(int n){
		double[] array = new double[n];
		for(int j =0 ;j<n;j++)
			array[j] = in.nextDouble();
		return array;
	}
	
	static int minIndex(int[] array){
		int minValue = Integer.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	
	static int minIndex(long[] array){
		long minValue = Long.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	
	static int minIndex(double[] array){
		double minValue = Double.MAX_VALUE;
		int minIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] < minValue){
				minValue = array[j];
				minIndex = j;
			}
		}
		return minIndex;
	}
	
	
	static int maxIndex(int[] array){
		int maxValue = Integer.MIN_VALUE;
		int maxIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] > maxValue){
				maxValue = array[j];
				maxIndex = j;
			}
		}
		return maxIndex;
	}
	
	static int maxIndex(long[] array){
		long maxValue = Long.MIN_VALUE;
		int maxIndex = -1;
		for(int j = 0;j<array.length;j++){
			if (array[j] > maxValue){
				maxValue = array[j];
				maxIndex = j;
			}
		}
		return maxIndex;
	}
	
	
	static long power(long x, long y){
		if (y == 0)
			return 1;
		if (y%2 == 1)
			return (x*power(x, y-1))%mod;
		return power((x*x)%mod, y/2)%mod;
	}
	static void verdict(boolean a){
        out.println(a ? "YES" : "NO");
    }
	
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } 
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try{
                str = br.readLine();
            } 
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
    }	
}