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.
# include <iostream>
# include <algorithm>
# include <string>
using namespace std;

int n;

int main()
{
    cin >> n;
    while(n--)
    {
        string ch;
        cin >> ch;
        int k = ch.size() - 1;
        int i = 0;
        for(; i <= k ; i++)
        {
            if(ch[i] == 'a')
            {
                break;
            }
        }
        if(ch[i] != 'a')
        {
            cout << "NO" << endl;
            continue;
        }
        int l = i,r = i;
        char t = 'a';
        bool flag = true;
        while(l > 0 || r < k)
        {
            if(l > 0 && ch[l - 1] == t + 1)
            {
                //printf("啦啦啦%c %c",ch[l - 1], t +1);
                l = l - 1;
                t = t + 1;
                continue;;
            }
            if(r < k && ch[r + 1] == t + 1)
            {
                r = r + 1;
                t = t + 1;
                continue;
            }
            flag = false;
            break;
        }
        if(flag)
        {
            cout << "YES" << endl;
        }
        else
        {
            cout << "NO" << endl;
        }
    }
    return 0;
}
