Methodius received an email from his friend Polycarp. However, Polycarp's keyboard is broken, so pressing a key on it once may cause the corresponding symbol to appear more than once (if you press a key on a regular keyboard, it prints exactly one symbol).

For example, as a result of typing the word "hello", the following words could be printed: "hello", "hhhhello", "hheeeellllooo", but the following could not be printed: "hell", "helo", "hhllllooo".

Note, that when you press a key, the corresponding symbol must appear (possibly, more than once). The keyboard is broken in a random manner, it means that pressing the same key you can get the different number of letters in the result.

For each word in the letter, Methodius has guessed what word Polycarp actually wanted to write, but he is not sure about it, so he asks you to help him.

You are given a list of pairs of words. For each pair, determine if the second word could be printed by typing the first one on Polycarp's keyboard.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of pairs to check. Further input contains n descriptions of pairs.

The first line of each description contains a single non-empty word s consisting of lowercase Latin letters. The second line of the description contains a single non-empty word t consisting of lowercase Latin letters. The lengths of both strings are not greater than 10^6.

It is guaranteed that the total length of all words s in the input is not greater than 10^6. Also, it is guaranteed that the total length of all words t in the input is not greater than 10^6.

Output

Output n lines. In the i-th line for the i-th pair of words s and t print YES if the word t could be printed by typing the word s. Otherwise, print NO.

Examples

Input


4
hello
hello
hello
helloo
hello
hlllloo
hello
helo


Output


YES
YES
NO
NO


Input


5
aa
bb
codeforces
codeforce
polycarp
poolycarpp
aaaa
aaaab
abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz


Output


NO
NO
YES
NO
NO
import java.util.*;
import java.io.*;

public class B {

	static final boolean stdin = true;
	static final String filename = "";
	static FastScanner br;
	static PrintWriter pw;

	public static void main(String[] args) throws IOException {

		if (stdin) {
			br = new FastScanner();
			pw = new PrintWriter(new OutputStreamWriter(System.out));
		} else {
			br = new FastScanner(filename + ".in");
			pw = new PrintWriter(new FileWriter(filename + ".out"));
		}

		Solver solver = new Solver();
		solver.solve(br, pw);
	}

	static class Solver {
		static long mod = 1000000000;

		public void solve(FastScanner br, PrintWriter pw) throws IOException {
			int n = br.nextInt();
			for(int r = 0; r < n; r++){
				String s1 = br.nextToken(), s2 = br.nextToken();
				ArrayList<Point1> a = new ArrayList<Point1>();
				ArrayList<Point1> b = new ArrayList<Point1>();
				a.add(new Point1(1, s1.charAt(0)));
				b.add(new Point1(1, s2.charAt(0)));
				for(int i = 1; i < s1.length(); i++){
					if(a.get(a.size()-1).b == s1.charAt(i)){
						a.set(a.size()-1, new Point1(a.get(a.size()-1).a+1, s1.charAt(i)));
					}else{
						a.add(new Point1(1, s1.charAt(i)));
					}
				}
				for(int i = 1; i < s2.length(); i++){
					if(b.get(b.size()-1).b == s2.charAt(i)){
						b.set(b.size()-1, new Point1(b.get(b.size()-1).a+1, s2.charAt(i)));
					}else{
						b.add(new Point1(1, s2.charAt(i)));
					}
				}
				if(a.size() != b.size()){
					pw.println("NO");
				}else{
					boolean done = false;
					for(int i = 0; i < a.size(); i++){
						if(a.get(i).b != b.get(i).b || a.get(i).a > b.get(i).a){
							pw.println("NO");
							done = true;
							break;							
						}
					}
					if(!done)
						pw.println("YES");
				}
			}
			pw.close();
		}

		static long gcd(long a, long b) {
			if (a > b)
				return gcd(b, a);
			if (a == 0)
				return b;
			return gcd(b % a, a);
		}

		static long pow(long a, long b) {
			if (b == 0)
				return 1L;
			long val = pow(a, b / 2);
			if (b % 2 == 0)
				return val * val % mod;
			else
				return val * val % mod * a % mod;
		}

	}

	static class Point1 implements Comparable<Point1> {
		int a;
		char b;

		Point1(int a, char b) {
			this.a = a;
			this.b = b;
		}

		@Override
		public int compareTo(Point1 o) {
			return this.a - o.a;
		}

	}

	static class Point2 implements Comparable<Point2> {
		int a;
		int b;

		Point2(int a, int b) {
			this.a = a;
			this.b = b;
		}

		@Override
		public int compareTo(Point2 o) {
			return this.b - o.b;
		}

	}

	static class UnionFind {
		// The number of elements in this union find
		private int size;
		// Used to track the size of each of the component
		private int[] sz;
		// id[i] points to the parent of i, if id[i] = i then i is a root node
		private int[] id;
		// Tracks the number of components in the union find
		private int numComponents;

		public UnionFind(int size) {
			if (size <= 0)
				throw new IllegalArgumentException("Size <= 0 is not allowed");
			this.size = numComponents = size;
			sz = new int[size];
			id = new int[size];
			for (int i = 0; i < size; i++) {
				id[i] = i; // Link to itself (self root)
				sz[i] = 1; // Each component is originally of size one
			}
		}

		// Find which component/set 'p' belongs to in amortized constant time.
		public int find(int p) {
			// Find the root of the component/set
			int root = p;
			while (root != id[root])
				root = id[root];
			// Compress the path leading back to the root.
			// Doing this operation is called "path compression"
			// and is what gives us amortized time complexity.
			while (p != root) {
				int next = id[p];
				id[p] = root;
				p = next;
			}
			return root;
		}

		// This is an alternative recursive formulation for the find method
		// public int find(int p) {
		// if (p == id[p]) return p;
		// return id[p] = find(id[p]);
		// }

		// Return whether or not the elements 'p' and
		// 'q' are in the same components/set.
		public boolean connected(int p, int q) {
			return find(p) == find(q);
		}

		// Return the size of the components/set 'p' belongs to
		public int componentSize(int p) {
			return sz[find(p)];
		}

		// Return the number of elements in this UnionFind/Disjoint set
		public int size() {
			return size;
		}

		// Returns the number of remaining components/sets
		public int components() {
			return numComponents;
		}

		// Unify the components/sets containing elements 'p' and 'q'
		public void unify(int p, int q) {
			int root1 = find(p);
			int root2 = find(q);
			// These elements are already in the same group!
			if (root1 == root2)
				return;
			// Merge smaller component/set into the larger one.
			if (sz[root1] < sz[root2]) {
				sz[root2] += sz[root1];
				id[root1] = root2;
			} else {
				sz[root1] += sz[root2];
				id[root2] = root1;
			}
			// Since the roots found are different we know that the
			// number of components/sets has decreased by one
			numComponents--;
		}
	}

	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(nextToken());
		}

		long nextLong() {
			return Long.parseLong(nextToken());
		}

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

		ArrayList<Integer>[] nextGraph(int nodes, int edges) {
			ArrayList<Integer>[] adj = new ArrayList[nodes];
			for (int i = 0; i < nodes; i++) {
				adj[i] = new ArrayList<Integer>();
			}
			for (int i = 0; i < edges; i++) {
				int a = nextInt() - 1;
				int b = nextInt() - 1;
				adj[a].add(b);
				adj[b].add(a);
			}
			return adj;
		}
	}
}