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.
for i in range(int(input())):
    word = input()
    ans = 'yes'
    if word[0] > word[-1]:
        s = 1
        f = len(word) - 1
    elif word[0] < word[-1]:
        s = 0
        f = len(word) - 2
    else:
        if len(word) > 1:
            ans = 'no'
        s = f = 0
    now = ord(max(word[0], word[-1]))
    while s < f:
        #print(ord(word[s]), word[s], ord(word[f]), word[f], now, )
        if f - s == 1 and not ((word[f] == 'a' and word[s] == 'b') or (word[f] == 'b' and word[s] == 'a')):
            f = s
            ans = 'no'
        if ord(word[s]) == now - 1:
            s += 1
            now -= 1
        elif ord(word[f]) == now - 1:
            f -= 1
            now -= 1
        else:
            s = f
            ans = 'no'
    if len(word) == 2 and (word != 'ab' and word != 'ba'):
        print('no')
    elif (word[s] == 'a' or word[f] == 'a'):
        print(ans)
    else:
        print('no')