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<cstring>

using namespace std;

const int N=30;
int cnt[N];

int main()
{

    int n;
    cin>>n;
    
    while(n--)
    {
        string res;
        cin>>res;
        int k=res.size();
        int ddd=0,xxx=100;
        int wz;
        for(int i=0;i<k;i++)
        {
            ddd=max(ddd,res[i]-'a'+1);
            xxx=min(xxx,res[i]-'a'+1);
            int e=i;
            
            cnt[++e]=res[i]-'a'+1;
            
            if((res[i]-'a'+1)==1)
            wz=i+1;
            
        }
        
        if((ddd-xxx+1)!=k||xxx!=1||ddd!=k)
        cout<<"NO"<<endl;
        else
        {
            int l,r;
            l=wz-1; 
            r=wz+1; 
            int ans=2;
            int flag=0;
            if(k==1)
            cout<<"YES"<<endl;
            else
            {
                 for(int i=2;i<=k;i++)
            {
               
                if(cnt[l]==ans||cnt[r]==ans)
                {
                    
                    if(cnt[l]==ans)
                    l=l-1;
                    else if(cnt[r]==ans)
                    r=r+1;
                    
                    ans++;
                }
                else
                {
                    flag=1;
                    cout<<"NO"<<endl;
                    break;
                }
                
            }
            
            if(flag==0)
            cout<<"YES"<<endl;
            }
           
        }
       
        
        memset(cnt,0,sizeof cnt);
       
    }
	
	return 0;
} 
