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.*;

public class B {

    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int T = fs.nextInt();
        outer: for (int tt = 0; tt < T; ++tt) {
            char[] c = fs.next().toCharArray();
            int[] counts=new int[26];
            for (char cc : c) counts[cc-'a']++;
            for (int i=0; i<26; i++) {
                if (counts[i]>1 || (i>0 &&counts[i]==1 && counts[i-1]==0)) {
                    System.out.println("NO");
                    continue outer;
                }
            }

            int n = c.length;
            int aIdx = 0;
            for(int i=0; i< c.length; ++i) if(c[i]=='a') aIdx = i;
            int l = aIdx, r = aIdx+1;
            int target = 'a';
            while(true) {
                boolean found = false;
                if(l >= 0 && c[l] == target) {
                    l--;
                    target++;
                    found = true;
                }
                if(r < n && c[r] == target) {
                    r++;
                    target++;
                    found = true;
                }
                if(!found) break;
            }
            System.out.println(l >0 || r < n ? "NO" : "YES");
        }
    }


    static final Random random=new Random();

    static void ruffleSort(int[] a) {
        int n=a.length;//shuffle, then sort
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }


}
