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.util.*;
	import java.io.*;

			public class codeforces
			{	
				public static long M = 1000000007;
				public static TreeSet<Integer> adj[];
			
				public static void main(String[] args)
				{
					InputReader in = new InputReader(System.in);
					PrintWriter pw = new PrintWriter(System.out);
					
					int T = in.nextInt();
					
					while(T-- > 0)
					{
						String s = in.nextLine();
						String t = in.nextLine();
						String p = in.nextLine();
						
						int j = 0;
						boolean flag = true;
						for(int i=0;i<s.length();i++)
						{
							char c = s.charAt(i);
							while(j < t.length() && t.charAt(j) != c)
								j++;
							if(j == t.length())
								flag = false;
							j++;
						}
						HashMap<Character, Integer> maps = new HashMap<>();
						HashMap<Character, Integer> mapt = new HashMap<>();
						HashMap<Character, Integer> mapp = new HashMap<>();
						for(int i=0;i<s.length();i++)
						{
							char c = s.charAt(i);
							if(!maps.containsKey(c))
								maps.put(c, 1);
							else
								maps.put(c, maps.get(c)+1);
						}
						for(int i=0;i<t.length();i++)
						{
							char c = t.charAt(i);
							if(!mapt.containsKey(c))
								mapt.put(c, 1);
							else
								mapt.put(c, mapt.get(c)+1);
						}
						for(int i=0;i<p.length();i++)
						{
							char c = p.charAt(i);
							if(!mapp.containsKey(c))
								mapp.put(c, 1);
							else
								mapp.put(c, mapp.get(c)+1);
						}
						for(Character c : maps.keySet())
						{
							if(!mapt.containsKey(c) || maps.get(c) > mapt.get(c))
							{
								flag = false;
								break;
							}
							else
							{
								mapt.put(c, mapt.get(c)-maps.get(c));
								if(mapt.get(c) == 0)
									mapt.remove(c);
							}
						}
						for(Character c : mapt.keySet())
						{
							if(!mapp.containsKey(c) || mapt.get(c) > mapp.get(c))
							{
								flag = false;
								break;
							}
						}
						if(flag)
							pw.println("YES");
						else
							pw.println("NO");
					}
					
					pw.flush();
					pw.close();
				}
				
				public static long[][] matrixpower(long[][] mat, long n, long p)
				{
					int r = mat.length;
					long[][] ans = new long[r][r];
					
					for(int i=0;i<r;i++)
						ans[i][i] = 1;
					
					while(n > 0)
					{
						if(n%2 == 1)
							ans = matrixmultiply(ans, mat, p);
						mat = matrixmultiply(mat, mat, p);
						n /= 2;
					}
					
					return ans;
				}
				
				public static long[][] matrixmultiply(long[][] M1, long[][] M2, long p)
				{
					int m1 = M1.length, m2 = M2[0].length;
					long[][] ans = new long[m1][m2];
					
					for(int i=0;i<m1;i++)
					{
						for(int j=0;j<m2;j++)
						{
							for(int k=0;k<M2.length;k++)
							{
								ans[i][j] += M1[i][k]*M2[k][j];
								if(ans[i][j] >= p-1)
									ans[i][j] %= p-1;
							}
						}
						//System.out.println(Arrays.toString(ans[i]));
					}
					return ans;
				}
				
				public static long prime(long nth) {
				    
				    long num, count, i;
				    num=1;
				    count=0;
				 
				    while (count < nth){
				      num=num+1;
				      for (i = 2; i <= num; i++){ //Here we will loop from 2 to num
				        if (num % i == 0) {
				          break;
				        }
				      }
				      if ( i == num){//if it is a prime number
				        count = count+1;
				      }
				    }
				    return num;
				  }
			  
				
				public static int[] suffle(int[] a,Random gen)
				{
					int n = a.length;
					for(int i=0;i<n;i++)
					{
						int ind = gen.nextInt(n-i)+i;
						int temp = a[ind];
						a[ind] = a[i];
						a[i] = temp;
					}
					return a;
				}
				
				public static long modularExponentiation(long x,long n,long M)
				{
				    long result=1;
				    while(n>0)
				    {
				        if(n % 2 ==1)
				            result=(result * x)%M;
				        x=(x*x)%M;
				        n=n/2;
				    }
				    return result;
				}
				
				public static long modInverse(long A,long M)
				{
				    return modularExponentiation(A,M-2,M);
				}
				
				public static int BinarySearch_l(long[] a, long k)
				{
					int n = a.length;
					int i = 0, j = n-1;
					int mid = 0;
					
					if(k - a[n-1] > 0)
						return n;
					else
					{
						while(j - i > 0)
						{
							mid = (i+j)/2;
							
							if(k - a[mid] > 0)
								i = mid + 1;
							else
								j = mid;
						}
					}
					
					return i;
				}
				
				public static int BinarySearch_r(long[] a, long k)
				{
					int n = a.length;
					int i = 0, j = n-1;
					int mid = 0;
					
					if(k - a[0] < 0)
						return -1;
					else
					{
						while(j - i > 0)
						{
							mid = (int)Math.ceil((double)(i+j)/2);
							
							if(k - a[mid] >= 0)
								i = mid;
							else
								j = mid - 1;
						}
					}
					
					return i;
				}
				public static long fact(long n, long M)
				{
					long ans = 1;
					
					for(long i=n;i>0;i--)
					{
						ans = (ans*i)%M;
					}
					
					return ans;
				}
				
			/*	public static void DFS(int sourse)
				{
					int count = 0;
					visited[sourse] = true;
					
					for(int u : adj[sourse])
					{
						if(!visited[u])
						{
							DFS(u);
						}
					}
					
					
				}*/
				
				public static ArrayList Divisors(int n)
				{
				    ArrayList<Integer> div = new ArrayList<>();
				    
				    for (int i=1; i<=Math.sqrt(n); i++)
				    {
				        if (n%i == 0)
				        {
				        	div.add(i);
				                
				            if(n/i != i)
				            	div.add(n/i);
				        }
				    }
				    return div;
				}
			
				public static long GCD(long a,long b)
				{
					if(b==0)
						return a;
					else
						return GCD(b,a%b);
				}
				
				public static long LCM(long a,long b)
				{
					return (a*b)/GCD(a, b);
				}
				
				static class pair implements Comparable<pair>
				{
					Integer x, y;
					pair(int x,int y)
					{
						this.x=x;
						this.y=y;
					}
					
					public int compareTo(pair o) {
						int result = x.compareTo(o.x);
						if(result==0)
							result = y.compareTo(o.y);
						
						return result;
					}  
					
					public String toString()
					{
						return x+" "+y;
					}
					
					public boolean equals(Object o)
					{
						if (o instanceof pair)
					    {
							pair p = (pair)o;
							return p.x - x == 0 && p.y - y == 0 ;
						}
						return false;
					}
					
					public int hashCode()
					{
						return new Long(x).hashCode()*31 + new Long(y).hashCode();
					}
				}
				
				static class InputReader 
				{
					 
					private final InputStream stream;
					private final byte[] buf = new byte[8192];
					private int curChar, snumChars;
					private SpaceCharFilter filter;
			 
					public InputReader(InputStream stream) 
					{
						this.stream = stream;
					}
			 
					public int snext() 
					{
						if (snumChars == -1)
							throw new InputMismatchException();
						if (curChar >= snumChars) 
						{
							curChar = 0;
							try 
							{
								snumChars = stream.read(buf);
							} 
							catch (IOException e) 
							{
								throw new InputMismatchException();
							}
							if (snumChars <= 0)
								return -1;
						}
						return buf[curChar++];
					}
			 
					public int nextInt() 
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						int sgn = 1;
						if (c == '-')
					    {
							sgn = -1;
							c = snext();
						}
						int res = 0;
						do 
						{
							if (c < '0' || c > '9')
								throw new InputMismatchException();
							res *= 10;
							res += c - '0';
							c = snext();
						} while (!isSpaceChar(c));
						return res * sgn;
					}
			 
					public long nextLong()
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						int sgn = 1;
						if (c == '-') 
						{
							sgn = -1;
							c = snext();
						}
						long res = 0;
						do 
						{
							if (c < '0' || c > '9')
								throw new InputMismatchException();
							res *= 10;
							res += c - '0';
							c = snext();
						} while (!isSpaceChar(c));
						return res * sgn;
					}
			 
					public int[] nextIntArray(int n) 
					{
						int a[] = new int[n];
						for (int i = 0; i < n; i++) 
						{
							a[i] = nextInt();
						}
						return a;
					}
			 
					public String readString()
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						StringBuilder res = new StringBuilder();
						do 
						{
							res.appendCodePoint(c);
							c = snext();
						} while (!isSpaceChar(c));
						return res.toString();
					}
			 
					public String nextLine() 
					{
						int c = snext();
						while (isSpaceChar(c))
							c = snext();
						StringBuilder res = new StringBuilder();
						do 
						{
							res.appendCodePoint(c);
							c = snext();
						} while (!isEndOfLine(c));
						return res.toString();
					}
			 
					public boolean isSpaceChar(int c) 
					{
						if (filter != null)
							return filter.isSpaceChar(c);
						return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
					}
			 
					private boolean isEndOfLine(int c) 
					{
						return c == '\n' || c == '\r' || c == -1;
					}
			 
					public interface SpaceCharFilter
				    {
						public boolean isSpaceChar(int ch);
					}
				}
				
				static class CodeX {
		        public static void sort(long arr[]) {
		            merge_sort(arr, 0, arr.length - 1);
		        }
		
		        private static void merge_sort(long A[], long start, long end) {
		            if (start < end) {
		                long mid = (start + end) / 2;
		                merge_sort(A, start, mid);
		                merge_sort(A, mid + 1, end);
		                merge(A, start, mid, end);
		            }
		
		        }
		
		        private static void merge(long A[], long start,long mid, long end) {
		            long p = start, q = mid + 1;
		            long Arr[] = new long[(int)(end - start + 1)];
		            long k = 0;
		
		            for (int i = (int)start; i <= end; i++) {
		                if (p > mid)
		                    Arr[(int)k++] = A[(int)q++];
		
		                else if (q > end)
		                    Arr[(int)k++] = A[(int)p++];
		
		                else if (A[(int)p] < A[(int)q])
		                    Arr[(int)k++] = A[(int)p++];
		
		                else
		                    Arr[(int)k++] = A[(int)q++];
		            }
		            for (int i = 0; i < k; i++) {
		                A[(int)start++] = Arr[i];
		            }
		
		        }
		
		    }
			}