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 ll long long
const ll N = 2e5 + 5;
ll n;
string s;
void solve(){
	cin >> s;
	int temp = 1;
	int i = 0, j = 0;
	for (i;i<s.size();i++){
		if (s[i] == 'a') break;
	}
	j = i - 1;
	i++;
	for (int t = 1; t < s.size();t++){
		if (i == s.size()){
			if (s[j] - 'a' != temp) break;
			j--;
		}
		else if (j == -1){
			if (s[i] - 'a' != temp) break;
			i++;
		}
		else if (s[i] - 'a' == temp)i++;
		else if (s[j] - 'a' == temp) j--;
		else break;
		temp++;
	}
	if (i == s.size()&& j == -1) cout << "YES\n";
	else cout << "NO\n";

}
int main(int argc, char const *argv[])
{
	int t; cin >> t;
	while (t--){
		solve();
	}
    return 0;  
}