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 _ in range(int(input())):
    string = input()
    arr = []
    flag = 0
    for i in string:
        if (ord(i)-96) in arr:
            flag = 1
            break
        arr.append(ord(i)-96)
    if flag == 1:
        print("NO")
        continue
    n = len(arr)
    
    # print(arr)

    if 1 not in arr:
        print("NO")
        continue

    if max(arr) != n:
        print("NO")
    
    else:
        idx = arr.index(1)
        for i in range(idx+1, n-1):
            if arr[i]>=arr[i+1]:
                flag = 1
                break
        for i in range(0, idx-1):
            if arr[i]<=arr[i+1]:
                flag = 1
                break
        if flag == 0:
            print("YES")
        else:
            print("NO")
    