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.
import sys
_DEBUG=False
if _DEBUG:
    fd = open("input.txt", "r")
    sys.stdin = fd
for t in range(int(sys.stdin.readline())):
    s = list(input())
    flag = False
    # print(s)
    prev = max(ord(s[0]), ord(s[-1]))
    if not prev - ord('a') + 1 == len(s):
        print('NO')
        continue
    if prev == ord(s[0]):
        s.pop(0)
    else:
        s.pop()
    while(s):
        x = max(ord(s[0]), ord(s[-1]))
        if not prev - x == 1:
            print("NO")
            flag = True
            break
        if x==ord(s[0]):
            s.pop(0)
        else:
            s.pop(-1)
        prev = x
    if flag:
        continue
    print("YES")
fd.close() if _DEBUG else None