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.
t=int(input())
for i in range(t):
    s=[]
    a=str(input())
    for i in a:
        s.append(ord(i))
    if len(set(s))!=len(s):
        print("NO")
    else:
        control=0
        if 97 in s:
            c=s.index(97)
        else:
            print("NO")
            continue
        x,y=1,1
        counter=98
        for i in range(len(a)-1):
            try:
                if s.index(counter)==c-x and c-x>=0:
                    counter += 1
                    x += 1
                    continue
                if s.index(counter)==c+y:
                    counter += 1
                    y += 1
                    continue
                if (s.index(counter)!=c-x and s.index(counter)!=c+y) and c-x>=0:
                    control+=1
                    break
                elif s.index(counter)!=c+y and c-x<0:
                    control += 1
                    break
            except ValueError:
                control+=1
                break
        if control!=0:
            print("NO")
        else:
            print("YES")