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.
import java.util.*;
import java.io.*;

public class Solution {

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader scan=new BufferedReader(new InputStreamReader(System.in));
		int t=Integer.parseInt(scan.readLine());
		while(t-->0) {
			String a = scan.readLine().trim();
			boolean flag=true;
			Deque<Character> dq=new ArrayDeque<>();
			char ch='a';
			
			for(int i=0;i<a.length();i++) {
				ch=(char) Math.max(ch,a.charAt(i));
				dq.add(a.charAt(i));
			}
			
			//System.out.println(ch);
					for(char c=ch;c>='a';c--) {
						if(dq.size()==0) {
							flag=false;
							break;
						}
						if(dq.peekFirst()==c) {
							dq.pollFirst();
						}else if(dq.peekLast()==c) {
							dq.pollLast();
						}else {
							flag=false;
							break;
						}
					}
					
					if(flag) {
						if(dq.size()>0) flag=false;
					}
					

			if(flag) {
				System.out.println("Yes");
			}else {
				System.out.println("No");
			}
			
		}
	}
	
}
