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())
c=list()
c.append('a')
c.append('b')
c.append('c')
c.append('d')
c.append('e')
c.append('f')
c.append('g')
c.append('h')
c.append('i')
c.append('j')
c.append('k')
c.append('l')
c.append('m')
c.append('n')
c.append('o')
c.append('p')
c.append('q')
c.append('r')
c.append('s')
c.append('t')
c.append('u')
c.append('v')
c.append('w')
c.append('x')
c.append('y')
c.append('z')
while t>0:
    t=t-1
    s=input()
    n=len(s)
    l=c[0:n]
    k=n-1
    for i in range(0,n):
        if s[0]==l[k]:
            s=s[1:len(s)]
            k=k-1
        elif s[-1]==l[k]:
            s=s[0:len(s)-1]
            k=k-1
        else:
            break
    if len(s)==0:
        print("YES")
    else:
        print("NO")