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.
#######puzzleVerma#######


import sys
import math
mod = 10**9+7


LI=lambda:[int(k) for k in input().split()]
input = lambda: sys.stdin.readline().rstrip()
IN=lambda:int(input())
S=lambda:input()
r=range

alpha="abcdefghijklmnopqrstuvwxyz"
for t in r(IN()):
    s=S()
    s=list(s)
    n=len(s)
    chk=[]
    for ele in s:
        if ele not in alpha[:n] or ele in chk:
            print("NO")
            break
        else:
            chk+=[ele]
    else:
        for i in range(1,n):
            it=alpha[i]
            fer=s.index(it)
            iit=alpha[i-1]
            ffer=s.index(iit)
            if abs(fer-ffer)>1:
                print("NO")
                break
            s.remove(iit)
        else:
            print("YES")
