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.
//clear adj and visited vector declared globally after each test case
//check for long long overflow
//while adding and subs check if mod becomes -ve
//while using an integer directly in a builtin function add ll
//Mod wale question mein last mein if dalo ie. Ans<0 then ans+=mod;
//Dont keep array name as size or any other key word
//ctrl+end - to break infinite loop in sublime text
 
#include <bits/stdc++.h>
#include <algorithm>
#include <cmath>
using namespace std;  
#define Fio                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(nullptr);                 \
    cout.tie(nullptr)
#define ll long long int
#define fo(i, a, b) for (long long int i = a; i <= b; i++)
#define w(t)  \
    int t;    \
    cin >> t; \
    while (t--)
#define all(v) v.begin(),v.end()
#define vi vector<int>
#define vl vector<long long int>
#define vvi vector<vector<int>>
#define vvl vector<vector<long long int>>
#define mii map<int, int>
#define umii unordered_map<int, int>
#define mll map<ll,ll>
#define umll unordered_map<ll,ll>
#define newl cout<<"\n"
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const ll inf = 1e9+7;
const ll mod = 998244353 ;
#define MAX 100005 

void call_priyanshu(){
    Fio;
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt", "w", stdout);
    #endif
}

ll mini(ll a,ll b){
    if(a>=b)return b;
    return a;
}
 
ll maxi(ll a,ll b){
    if(a>=b)return a;
    return b;
}

void solve(){
    //write your code here
    string s;
        cin>>s;
        ll n = s.length();
        string ans = "YES";
        vl p(27,0);
        for(int i = 0;i<n;i++){
            if(p[s[i]-'a']!=0)ans="NO";
            p[s[i]-'a']=i+1;
        }
        ll m = p[0];
        ll l = m,h=m;
        ll fr=0;
        if(m==0){
            fr=1;
        }
        fo(i,1,25){
            if(p[i]==0){
                fr=1;
                continue;
            }
            if(fr)ans="NO";
            if(p[i]==l-1){
                l--;
                continue;
            }
            else if(p[i]==h+1){
                h++;
                continue;
            }
            ans ="NO";
        }
        cout<<ans;
        newl;
}

int main(){
    call_priyanshu();
    w(t){
      solve();
    }
    return 0;
}