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>

using namespace std;

#define fio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define PI 3.14159265358979323846264338327950l;
ll gcd(ll a, ll b){if(b==0) return a; return gcd(b,a%b);}
ll lcm(ll a, ll b){return (a*b)/gcd(a,b);}
ll power(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
ll ncp(ll n, ll p){if(p==0){return 1;}ll ans=1ll;for(ll i=n;i>n-p;i--){ans*=i;} for(ll i=1;i<=p;i++){ans/=i;} return ans;}
 
ll problem(int x, int y,ll c)
{
	if(x==0)
	{
		return 0 ;
	} 
	
	if(y==0)
	{
		return 1 ;
	}
	
	ll p ;
	
	if(y%2==0)
	{
		p= problem(x,y/2,c) ;
		p= (p*p)%c ;
	}
	
	if(y%2==1)
	{
		p = x%c ;
		
		p =(p*problem(x,y-1,c)%c)%c ;
	}
	
	 return (ll)((p + c) % c);
}
 
int main()
{
    int t;
    cin >> t;
    for(int o=0;o<t;o++){
    	string s;
    	cin >> s;
        int count=96+s.length(),k=1;
        while(s.length()){
        	if(s[0]==char(count))
        	s=s.substr(1,s.length()-1);
        	else if(s[s.length()-1]==char(count))
        	s=s.substr(0,s.length()-1);
        	else{
        		k=0;
        		cout << "NO\n";
        		break;
        	}
        	count--;
        }
        if(k)
        cout << "YES\n";
    }
    
    return 0;
}