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 <bits/stdc++.h>
#include <stdlib.h>
using namespace std ;
#define int long long
#define pb push_back
#define ff first
#define ss second
#define inf 2e18
#define poke ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL) ;
int32_t main() {
    poke;
    int t;
    cin >> t ;
    while(t -- ){
        string s ;
        cin >> s ;
        int n = s.length() ;
        int l =-1,r =-1;
        for(int i = 0; i <n ;i++){
            if(s[i] == 'a'){
                l = i,  r = i ;
            }
        }
        if(l == -1){
            cout << "NO\n";
            continue ;
        }
        string res = "YES" ;
        int cnter = 1;
        while(cnter < n){
            bool f = false ;
            if( (l-1 >= 0 && s[l-1] == ('a'+cnter) ) ){
                f = true ;
                l--;
            }
            if( (r+1 < n && s[r+1] == ('a'+cnter) ) ){
                f = true ;
                r++;
            }
            if(!f){
                res = "NO" ;
                break ;
            }
            cnter ++ ;
        }
        cout << res << "\n" ;
    }
    return 0 ;
}
