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>
#define fast ios_base::sync_with_stdio(0);ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ll long long
#define ull unsigned long long
#define ld long double
#define endl "\n"
#define pb push_back
#define F first
#define S second
#define Yes cout << "YES" << endl
#define No cout << "NO" << endl
using namespace std;

const int N = 2e5 + 7;
const int mod = 1e9 + 7;
string s ;
char ch[] = {'a' , 'b' , 'c' , 'd' , 'e' , 'f' , 'g' , 'h' , 'i' , 'j' , 'k' , 'l' , 'm' , 'n' , 'o' , 'p' , 'q' , 'r' , 's' , 't' , 'u' , 'v' , 'w' , 'x' , 'y' , 'z' } ;

void solve()
{
    cin >> s ;
    ll n = s.size(), q = 0;
    for(int i = 0 ; i < n; i++)
    {
        if(s[i] == 'a')
            q = i;
    }
    bool f1 = 0, f2 = 0, is = 0;
    ll l = q, r = q + 1, cur = 0;
    while(l >= 0 || r < n)
    {
        if(l >= 0 && s[l] == ch[cur]) l-- , cur ++;
        else if(r < n && s[r] == ch[cur]) r++, cur++;
        else f1 = 1 , f2 = 1;
        if(f2 && f1)
        {
            is = 1;
            break;
        }
    }
    if(f2 && f1) is = 1;
    if(!is) Yes;
    else No;
}

int main()
{
    fast;
    int t = 1;
    cin >> t;
    while (t--){
        solve();
    }
    return 0;
}
