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 xr in range(t):
    s = input()
    n = len(s)
    i1 = -1
    i2 = -1
    us = [0] * 26
    k = [0] * 26
    for i in range(len(s)):
        if s[i] == 'a':
            if i1 != -1:
                i2 = i
            else:
                i1 = i
        us[ord(s[i]) - ord('a')]+=1
        k[ord(s[i]) - 97] = i
    if i2 != -1 or i1 == -1:
        print('NO')
        continue
    i2 = i1
    x = 1
    f = 1
    for i in range(1, 26):
        if us[i] == 0:
            continue
        if us[i] != 0 and us[i - 1] == 0:
            print("NO")
            f = 0
            break
        elif us[i] > 1:
            print("NO")
            f = 0
            break
        t = chr(97 + i)
        if k[i] == i1 - 1:
            i1 -= 1
        elif k[i] == i2 + 1:
            i2 += 1
        else:
            print("NO")
            f = 0
            break
    if f:
        print("YES")
