Polycarp wants to assemble his own keyboard. Layouts with multiple rows are too complicated for him — his keyboard will consist of only one row, where all $26$ lowercase Latin letters will be arranged in some order.

Polycarp uses the same password $s$ on all websites where he is registered (it is bad, but he doesn't care). He wants to assemble a keyboard that will allow to type this password very easily. He doesn't like to move his fingers while typing the password, so, for each pair of adjacent characters in $s$, they should be adjacent on the keyboard. For example, if the password is abacaba, then the layout cabdefghi... is perfect, since characters a and c are adjacent on the keyboard, and a and b are adjacent on the keyboard. It is guaranteed that there are no two adjacent equal characters in $s$, so, for example, the password cannot be password (two characters s are adjacent).

Can you help Polycarp with choosing the perfect layout of the keyboard, if it is possible?


-----Input-----

The first line contains one integer $T$ ($1 \le T \le 1000$) — the number of test cases.

Then $T$ lines follow, each containing one string $s$ ($1 \le |s| \le 200$) representing the test case. $s$ consists of lowercase Latin letters only. There are no two adjacent equal characters in $s$.


-----Output-----

For each test case, do the following:

  if it is impossible to assemble a perfect keyboard, print NO (in upper case, it matters in this problem);  otherwise, print YES (in upper case), and then a string consisting of $26$ lowercase Latin letters — the perfect layout. Each Latin letter should appear in this string exactly once. If there are multiple answers, print any of them. 


-----Example-----
Input
5
ababa
codedoca
abcda
zxzytyz
abcdefghijklmnopqrstuvwxyza

Output
YES
bacdefghijklmnopqrstuvwxyz
YES
edocabfghijklmnpqrstuvwxyz
NO
YES
xzytabcdefghijklmnopqrsuvw
NO

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "for _ in range(int(input())):\n    q=input()\n    ans=q[0]\n    test=[0]*26\n    j=0\n    c = 1\n    for i in q[1:]:\n        if j>0 and ans[j-1]==i:\n            j-=1\n            continue\n        if j<len(ans)-1 and ans[j+1]==i:\n            j+=1\n            continue\n        if j==0:\n            ans=i+ans\n            continue\n        if j==len(ans)-1:\n            ans+=i\n            j+=1\n            continue\n        c=0\n    for i in ans:test[ord(i)-97]+=1\n    for i in range(26):\n        if test[i]>1:c=0\n        if test[i]==0:ans+=chr(i+97)\n    if c:\n        print('YES')\n        print(ans)\n    else:\n        print('NO')"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13351