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.
//package codeforces.round731div3;

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

import static java.lang.Math.*;

//Think through the entire logic before jump into coding!
//If you are out of ideas, take a guess! It is better than doing nothing!
//Read both C and D, it is possible that D is easier than C for you!

//Be aware of integer overflow!
//If you find an answer and want to return immediately, don't forget to flush before return!

public class B {
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        //initReaderPrinter(true);
        initReaderPrinter(false);
        solve(in.nextInt());
        //solve(1);        
    }

    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            String s = in.next();
            int i = 0;
            for(; i < s.length(); i++) {
                if(s.charAt(i) == 'a') break;
            }
            boolean ans = i < s.length();
            if(ans) {
                int l = i - 1, r = i + 1;
                char c = 'b';
                while(l >= 0 || r < s.length()) {
                    if(l >= 0 && s.charAt(l) == c) {
                        l--;
                    }
                    else if(r < s.length() && s.charAt(r) == c) {
                        r++;
                    }
                    else {
                        ans = false;
                        break;
                    }
                    c++;
                }
            }
            out.println(ans ? "YES" : "NO");
        }
        out.close();
    }

    static void initReaderPrinter(boolean test) {
        if (test) {
            try {
                in = new InputReader(new FileInputStream("src/input.in"));
                out = new PrintWriter(new FileOutputStream("src/output.out"));
            } catch (IOException e) {
                e.printStackTrace();
            }
        } else {
            in = new InputReader(System.in);
            out = new PrintWriter(System.out);
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        InputReader(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream), 32768);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        Integer[] nextIntArray(int n) {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitive(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        int[] nextIntArrayPrimitiveOneIndexed(int n) {
            int[] a = new int[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextInt();
            return a;
        }

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitive(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        long[] nextLongArrayPrimitiveOneIndexed(int n) {
            long[] a = new long[n + 1];
            for (int i = 1; i <= n; i++) a[i] = nextLong();
            return a;
        }

        String[] nextStringArray(int n) {
            String[] g = new String[n];
            for (int i = 0; i < n; i++) g[i] = next();
            return g;
        }
    }
}