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=input()
    d= {'a' : 1,'b' : 2, 
        'c' : 3, 
        'd' : 4, 
        'e' : 5, 
        'f' : 6, 
        'g' : 7, 
        'h' : 8, 
        'i' : 9, 
        'j' : 10, 
        'k' : 11, 
        'l' : 12, 
        'm' : 13, 
        'n' : 14, 
        'o' : 15, 
        'p' : 16, 
        'q' : 17, 
        'r' : 18, 
        's' : 19, 
        't' : 20, 
        'u' : 21, 
        'v' : 22, 
        'w' : 23, 
        'x' : 24, 
        'y' : 25, 
        'z' : 26
    }
    l=0
    r=len(s)-1
    flag=0
    d1={}
    for j in s:
        if j in d1:
            print("NO")
            flag=12
            break
        else:
            d1[j]=1
    if flag ==12:
        continue
    for n in range(len(s)):
        if d[s[l]]<d[s[r]]:
            if d[s[r]]-1==d[s[r-1]]:
                r=r-1
            elif d[s[r]]-1==d[s[l]]:
                r=r-1
            else:
                print("NO")
                flag=1
                break
        elif d[s[l]]>d[s[r]]:
            if d[s[l]]-1==d[s[l+1]]:
                l=l+1
            elif d[s[l]]-1==d[s[r]]:
                l=l+1
            else:
                print("NO")
                flag=1
                break
    if flag==0:
        if "a" in s:
            print("YES")
        else:
            print("NO")