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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;

public class R731_B {
    public static Scanner reader = new Scanner(System.in);
    
    // solution here.
    public static void solve() throws IOException
    {
    	String s = reader.nextLine();
    	int l = 0;
    	int h = s.length()-1;
    	int flag = 0;
    	int prev = Math.max(s.charAt(0), s.charAt(s.length()-1))+1;
    	int current = -1;
    	while(l<=h)
    	{
    		if(l == h)
    		{
    			if(s.charAt(l)!='a')
    				flag=1;
    		}
    		if(s.charAt(l)>s.charAt(h)) {
    			current = s.charAt(l);
    			l++;
    		}
    		else {
    			current = s.charAt(h);
    			h--;
    		}
    		if(current+1 != prev)
    		{
    			flag=1;
    			break;
    		}
    		else
    		{
    			prev = current;
    		}
    	}
    	if(flag == 1)
    		System.out.println("NO");
    	else
    		System.out.println("YES");
    	
    }
    
    public static void main(String[] args) {
        
        final int cases;
        try {
            cases = Integer.parseInt(reader.nextLine().trim());
            for (int i = 0; i < cases; i++) {
                solve();
            }

        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}