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.
l=["a", "b", "c", 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
for i in range(int(input())):
    s=list(str(input()))
    i=0
    n=len(s)
    k=0
    j=0
    
    while(j<n and s[j]!=l[k]):
        j+=1
        
    if(j==n):
        print("NO")
    else:
        k+=1
        if(j==n-1):
            i=n-2
            while(i>=0):
                if(s[i]==l[k]):
                    i-=1
                    k+=1
                else:
                    break
            if(i==-1):
                print("YES")
            else:
                print("NO")
        elif(j==0):
            i=1
            while(i<n):
                if(s[i]==l[k]):
                    i+=1
                    k+=1
                else:
                    break
            if(i==n):
                print("YES")
            else:
                print("NO")
        else:
            i=j-1
            j+=1
            c=0
            while(i>=0 and j<n):
                if(s[i]==l[k]):
                    i-=1
                    k+=1
                elif(s[j]==l[k]):
                    j+=1
                    k+=1
                else:
                    c=1
                    break
            while(i>=0 and c==0):
                if(s[i]==l[k]):
                    i-=1
                    k+=1
                else:
                    break
            while(j<n and c==0):
                if(s[j]==l[k]):
                    j+=1
                    k+=1
                else:
                    break
                
            if(i==-1 and j==n):
                print("YES")
            else:
                print("NO")
        
        
    



        
