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;
typedef long long ll;
typedef pair<int,int>pii;
typedef pair<long long,long long>pll;
void solve(){
    string s;
    char c='a';
    int n,i,p=100,l,r;
    cin>>s;
    n=s.length();
    for(i=0;i<n;i++){
        if(s[i]=='a') p=i;
    }
    if(p==100){
        printf("NO\n");
        return;
    }
    l=p-1,r=p+1;
    c=c+1;
    for(i=1;i<=n;i++){
        if(l>=0&&s[l]==c) l--,c=c+1;
        else if(r<n&&s[r]==c) r++,c=c+1;
    }
    if(c=='a'+n) printf("YES\n");
    else printf("NO\n");
}
int main(void){
    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
}