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
#define inf 0x7fffffff
#define nl '\n'
#define ml ' '
#define mp make_pair
#define N 200010
#define rip(m,n) for(int i=m;i<=n;i++)
const int mod = 998244353;
using namespace std;
void solve()
{
	string s;
	cin >> s;
	int mi = s.length() - 1;
	map<char, int>kk;
	rip(0, mi) {
		kk[s[i]]++;
		if (kk[s[i]] > 1) {
			cout << "no" << nl;
			return;
		}
	}
	deque<char>q;
	rip(0, mi)q.push_back(s[i]);
	while (!q.empty())
	{
		if (q.front() > q.back()) {
			char x = q.front(); q.pop_front();
			if (x - 1 == q.front() || x - 1 == q.back()) {}
			else {
				cout << "no" << nl;
				return;
			}
		}
		else {
			if (q.size() == 1&&q.front()=='a') {
				cout << "yes" << nl;
				return;
			}
			else if (q.size() == 1) {
				cout << "no" << nl;
				return;
			}
			char x = q.back(); q.pop_back();
			if (x - 1 == q.front() || x - 1 == q.back()) {
				continue;
			}
			else {
				cout << "no" << nl;
				return;
			}
		}
	}
	cout << "yes" << nl;
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);
	int t ;
	cin >> t;
	while (t--)solve();
	return 0;
}
//Thanks for being accepted!!!