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 ques2():
    s = input()
    index = -1 
    for i in range(len(s)):
        if s[i]=='a':
            index = i 
            break
        
    ans = 0
    if index == -1:
        ans = 1
    else:
        value = ord(s[index])
        l = index
        r = index 
        visited_coor = 1 
        while visited_coor < len(s):
            if l-1 >= 0 and ord(s[l-1]) == value+1:
                visited_coor += 1 
                l -= 1 
                value += 1 
            elif r+1 < len(s) and ord(s[r+1])==value+1:
                visited_coor += 1 
                r += 1 
                value += 1 
            else:
                break
            
        if visited_coor<len(s):
            ans=1
    if ans==0:
        return "YES"
    return "NO"
    
    
number_of_testcases = int(input())
for _ in range(number_of_testcases):
    print(ques2())