A string s of length n (1 ≤ n ≤ 26) is called alphabetical if it can be obtained using the following algorithm:

  * first, write an empty string to s (i.e. perform the assignment s := ""); 
  * then perform the next step n times; 
  * at the i-th step take i-th lowercase letter of the Latin alphabet and write it either to the left of the string s or to the right of the string s (i.e. perform the assignment s := c+s or s := s+c, where c is the i-th letter of the Latin alphabet). 



In other words, iterate over the n first letters of the Latin alphabet starting from 'a' and etc. Each time we prepend a letter to the left of the string s or append a letter to the right of the string s. Strings that can be obtained in that way are alphabetical.

For example, the following strings are alphabetical: "a", "ba", "ab", "bac" and "ihfcbadeg". The following strings are not alphabetical: "z", "aa", "ca", "acb", "xyz" and "ddcba".

From the given string, determine if it is alphabetical.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

Each test case is written on a separate line that contains one string s. String s consists of lowercase letters of the Latin alphabet and has a length between 1 and 26, inclusive.

Output

Output t lines, each of them must contain the answer to the corresponding test case. Output YES if the given string s is alphabetical and NO otherwise.

You can output YES and NO in any case (for example, strings yEs, yes, Yes and YES will be recognized as a positive answer).

Example

Input


11
a
ba
ab
bac
ihfcbadeg
z
aa
ca
acb
xyz
ddcba


Output


YES
YES
YES
YES
YES
NO
NO
NO
NO
NO
NO

Note

The example contains test cases from the main part of the condition.
answers = []

def solve(string):
  length = len(string)

  if length == 0:
    answers.append('YES')
    return
  
  newStore = {}
  for ele in string:
    if ele not in newStore:
      newStore[ele] = 1
    else:
      answers.append('NO')
      return

  order = 97
  temp = 0
  store = {}

  while temp != length:
    store[chr(order)] = 1
    temp += 1
    order += 1
  
  for ele in string:
    if ele not in store:
      answers.append('NO')
      return
  
  final = list(string)

  current = 97
  temp = 1

  limit = current + length - 1

  while current != limit:
    index = final.index(chr(current))

    prev = index-1
    after = index+1
    req = chr(current+1)

    if after != len(final):
      if final[after] == req:
        final.pop(index)
        current += 1
      else:
        if prev != -1:
          if final[prev] == req:
            final.pop(index)
            current += 1
          else:
            answers.append('NO')
            return
        else:
            answers.append('NO')
            return
    elif prev != -1:
      if final[prev] == req:
        final.pop(index)
        current += 1
      else:
        answers.append('NO')
        return
    else:
      answers.append('NO')
      return
  
  answers.append('YES')




T = int(input())
while T:
  s = input()
  solve(s)
  T -= 1

for ans in answers:
  print(ans)