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.
def solve(s):
  if len(s) > 26:
    return "No"
  l = 0
  r = len(s)-1
  lastrem = chr(len(s)+97)
  while l < r:
    if ord(s[l]) < ord(s[r]):
      if ord(lastrem) == ord(s[r]) + 1:
        lastrem = s[r]
        r -= 1
      else:
        return "No"
    elif ord(s[l]) > ord(s[r]):
      if ord(lastrem) == ord(s[l]) + 1:
        lastrem = s[l]
        l += 1
      else:
        return "No"
    else:
      return "No"
  if s[l] == 'a' or s[r] == 'a':
    return "Yes"
  else:
    return "No"

t = int(input())

for case in range(t):
  s = input()
  print(solve(s))