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.
for _ in range(int(input())):
	n=input()
	a=0
	b=len(n)-1
	m=chr(ord(max(n))+1)
	if b+1==2 or len(set(n))!=b+1:
		if n=='ab' or n=='ba':
			print("YES")
			continue
		else:
			print("NO")
			continue
	while a<=b:
		p=max(n[a],n[b])
		if  p<m and abs(ord(p)-ord(m))==1:
			m=p
			a+=(m==n[a])
			b-=(m==n[b])
		else:
			print("NO")
			break
			continue
	else:
		print("YES"*('a' in n)+"NO"*("a" not in n))
		continue
#11
#a
#ba
#ab
#bac
#ihfcbadeg
#z
#aa
#ca
#acb
#xyz
#ddcba
#abs(ord(m)-ord(p))>10