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>
#define ll long long
using namespace std;
int t;
char s[30];
int main()
{
	ios::sync_with_stdio(0);
	cin>>t;
	while(t--)
	{
	
	cin>>s+1;
	int len=strlen(s+1);
	int l=0,r=0,pd=1;
	for(int i=1;i<=len;i++)
	{
		if(s[i]=='a')
		{
			l=i;r=i;break;
		}
	}
	if(len==1)
	{
		if(s[1]=='a')
		{
			printf("YES\n");
		}
		else printf("NO\n");
		continue;
	}
	if(l==0)
	{
		printf("NO\n");continue;
	}
	else
	{
		
		for(int i=1;i<len&&pd==1;i++)
		{
			if(s[l-1]=='a'+i)l--;
			else if(s[r+1]=='a'+i)r++;
			else pd=0;
			if(pd==0)break;
		}
	}
	if(pd==1)
	{
		printf("YES\n");	
	}
	else 
	{
		printf("NO\n");	
	} 
	}
	return 0; 
}