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
#!python3
"""
Author: w1ld [at] inbox [dot] ru
"""

from collections import deque, Counter
import array
from itertools import combinations, permutations
from math import sqrt
# import unittest


def read_int():
    return int(input().strip())


def read_int_array():
    return [int(i) for i in input().strip().split(' ')]

######################################################

class Node:
    def __init__(self, c):
        self.c = c
        self.l = None
        self.r = None


tests = read_int()

for test in range(tests):
    s = input().strip()
    left = Node(s[0])
    x = left
    found = True
    used = set([x.c])
    for c in s[1:]:
        if x.c == c:
            continue
        if x.l and x.l.c == c:
            x = x.l
        elif x.r and x.r.c == c:
            x = x.r
        elif not x.l and c not in used:
            x.l = Node(c)
            used.add(c)
            x.l.r = x
            x = x.l
            left = x
        elif not x.r and c not in used:
            x.r = Node(c)
            used.add(c)
            x.r.l = x
            x = x.r
        else:
            found = False
            break

    if not found:
        print("NO")
    else:
        ans = []
        x = left
        while x:
            ans.append(x.c)
            x = x.r
        for c in 'abcdefghijklmnopqrstuvwxyz':
            if c not in used:
                ans.append(c)
        print("YES")
        print(''.join(ans))











#!python3
"""
Author: w1ld [at] inbox [dot] ru
"""

from collections import deque, Counter
import array
from itertools import combinations, permutations
from math import sqrt
# import unittest


def read_int():
    return int(input().strip())


def read_int_array():
    return [int(i) for i in input().strip().split(' ')]

######################################################

class Node:
    def __init__(self, c):
        self.c = c
        self.l = None
        self.r = None


tests = read_int()

for test in range(tests):
    s = input().strip()
    left = Node(s[0])
    x = left
    found = True
    used = set([x.c])
    for c in s[1:]:
        if x.c == c:
            continue
        if x.l and x.l.c == c:
            x = x.l
        elif x.r and x.r.c == c:
            x = x.r
        elif not x.l and c not in used:
            x.l = Node(c)
            used.add(c)
            x.l.r = x
            x = x.l
            left = x
        elif not x.r and c not in used:
            x.r = Node(c)
            used.add(c)
            x.r.l = x
            x = x.r
        else:
            found = False
            break

    if not found:
        print("NO")
    else:
        ans = []
        x = left
        while x:
            ans.append(x.c)
            x = x.r
        for c in 'abcdefghijklmnopqrstuvwxyz':
            if c not in used:
                ans.append(c)
        print("YES")
        print(''.join(ans))










