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.
#!/usr/bin/env python
# coding: utf-8

# In[27]:


T = int(input())
out = []
for t in range(T):
    s = input()
    n = len(s)
    mapping = dict()
    t = 'abcdefghijklmnopqrstuvwxyz'
    flag = 1
    for e in t:
        mapping[e] = -1
    
    for i in range(n):
        if mapping[s[i]] >=0:
            out.append('No')
            flag = 0
            break
        else:
            mapping[s[i]] = i
    
    if flag:
        for i in range(n):
            if mapping[t[i]] ==-1:
                flag= 0
                out.append('No')
                break
    if flag:   
        l = mapping['a'] -1
        r = mapping['a'] +1

        x = ord('b')
        while x < (ord('a') + n):
            #print(x)
            if l >=0 and s[l]== chr(x):
                x+=1
                l-=1
            elif r <n and s[r] == chr(x):
                x+=1
                r+=1
            else:
                out.append('No')
                flag = 0
                break
    if flag:
        out.append('Yes')


# In[28]:


for e in out:
    print(e)


# In[ ]:




