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>
typedef long long ll;
#define vll vector<ll>
#define pb push_back
#define pll pair<ll,ll>
#define vpll vector<pair<ll,ll> >
#define st(x) sort(x.begin(),x.end())
#define sz(x) (ll)x.size()
#define input(n,x) for(ll i=0;i<n;i++){ll k;cin>>k;x.pb(k);}
#define fr first
#define sc second
#define rep(i,j,k) for(ll i=j;i<k;i++)
#define repn(i,j,k) for(ll i=j;i>=k;i--)
using namespace std;
vll x;

bool solve(){
    string s;
    cin>>s;
    ll k = sz(s)-1;
    ll  st = 0,en = sz(s)-1;
    while(st<=en){
        if((s[st]-'a') == k){
            k--;
            st++;
        }
        else if((s[en]-'a') == k){
            k--;
            en--;
        }
        else{
            return false;
        }
    }
    return true;
}


int main() {
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	ll t;
	cin>>t;
	while(t--){
	    if(solve()){
	        cout<<"YES\n";
	    }
	    else{
	        cout<<"NO\n";
	    }
	}
	return 0;
}