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.
/*
Sunday 11 July 2021 10:45:54 AM IST
*/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;


void solve(){
	string s;
	cin >> s;
	int n = s.size();
	char c;
	int st = 0;
	int en = n - 1;
	c = max(s[st], s[en]);
	while(st < en){
		if(s[st] == c){
			st++;
		}
		else if(s[en] == c){
			en--;
		}
		else{
			cout << "NO\n";
			return;
		}
		c--;
	}
	if(c == 'a' && s[st] == 'a')
		cout << "YES\n";
	else
		cout << "NO\n";
	
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t = 1;
	cin >> t;
	while(t--)
		solve();
}
