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 PYTHON3):

Solution:
from sys import stdin, stdout
from collections import Counter

def rsingle_int():
    return int(stdin.readline().rstrip())
 
def rmult_int():
    return [ int(x) for x in stdin.readline().rstrip().split() ]
 
def rmult_str():
    return stdin.readline().rstrip().split()
 
def r_str():
    return stdin.readline().rstrip()
    
def rsingle_char():
    return stdin.read(1)
 
def eq_tile(one, two):
    return one[0] == two[0] and one[1] == two[1]

def is_substring(one, main):
    i, j = 0, 0
    one_len = len(one)
    while i < one_len and j < len(main):
        if one[i] == main[j]:
            i += 1
            j += 1
        else:
            j += 1
    return i == one_len

def main():
    q = rsingle_int()
    for i in range(q):
        s = r_str()
        t = r_str()
        p = r_str()
        
        if is_substring(s, t):
            s_letter_cnt = Counter(s)
            t_letter_cnt = Counter(t)
            p_letter_cnt = Counter(p)
            t_letter_cnt.subtract(s_letter_cnt)
            t_letter_cnt.subtract(p_letter_cnt)
            
            not_possible = False
            for key in t_letter_cnt.keys():
                if t_letter_cnt[key] > 0:
                    not_possible = True
                    break
            if not_possible:
                print("NO")
            else:
                print("YES")
        else:
            print("NO")
    
main();