Problem: You are given a string, consisting of lowercase Latin letters.

A pair of neighbouring letters in a string is considered ugly if these letters are also neighbouring in a alphabet. For example, string "abaca" contains ugly pairs at positions (1, 2) — "ab" and (2, 3) — "ba". Letters 'a' and 'z' aren't considered neighbouring in a alphabet.

Can you rearrange the letters of a given string so that there are no ugly pairs? You can choose any order of the letters of the given string but you can't add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

You also have to answer T separate queries.

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of queries.

Each of the next T lines contains string s (1 ≤ |s| ≤ 100) — the string for the next query. It is guaranteed that it contains only lowercase Latin letters.

Note that in hacks you have to set T = 1.

Output

Print T lines. The i-th line should contain the answer to the i-th query.

If the answer for the i-th query exists, then print such a rearrangment of letters of the given string that it contains no ugly pairs. You can choose any order of the letters of the given string but you can't add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

Otherwise print "No answer" for that query.

Example

Input


4
abcd
gg
codeforces
abaca


Output


cadb
gg
codfoerces
No answer

Note

In the first example answer "bdac" is also correct.

The second example showcases the fact that only neighbouring in alphabet letters are not allowed. The same letter is ok.

There are lots of valid answers for the third example.

Solution (in PYTHON3):

Solution:
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
import inspect, re


def varname(p):  # prints name of the variable
    for line in inspect.getframeinfo(inspect.currentframe().f_back)[3]:
        m = re.search(r'\bvarname\s*\(\s*([A-Za-z_][A-Za-z0-9_]*)\s*\)', line)
        if m:
            return m.group(1)


BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def lcm(a, b):
    return (a * b) / gcd(a, b)


mod = int(1e9) + 7


def power(k, n):
    if n == 0:
        return 1
    if n % 2:
        return (power(k, n - 1) * k) % mod
    t = power(k, n // 2)
    return (t * t) % mod


def totalPrimeFactors(n):
    count = 0
    if (n % 2) == 0:
        count += 1
        while (n % 2) == 0:
            n //= 2

    i = 3
    while i * i <= n:
        if (n % i) == 0:
            count += 1
            while (n % i) == 0:
                n //= i
        i += 2
    if n > 2:
        count += 1
    return count


# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
#     spf[1] = 1
#     for i in range(2, MAXN):
#         spf[i] = i
#     for i in range(4, MAXN, 2):
#         spf[i] = 2
#
#     for i in range(3, mt.ceil(mt.sqrt(MAXN))):
#         if (spf[i] == i):
#             for j in range(i * i, MAXN, i):
#                 if (spf[j] == j):
#                     spf[j] = i
#
#
# def getFactorization(x):
#     ret = 0
#     while (x != 1):
#         k = spf[x]
#         ret += 1
#         # ret.add(spf[x])
#         while x % k == 0:
#             x //= k
#
#     return ret


# Driver code

# precalculating Smallest Prime Factor
# sieve()


def answer_bata_lavde(a, b):
    for i in a:
        print(i, end='')
    for i in b:
        print(i, end='')
    print('')


def main():
    for _ in range(int(input())):
        S=input()
        s=[]
        o=[]
        e=[]
        for i in S:
            if ord('a')<=ord(i)<=ord('z'):
                if ord(i)%2:
                    o.append(i)
                else:
                    e.append(i)
        o.sort()
        e.sort()
        if len(o)==0:
            answer_bata_lavde([], e)
        elif len(e)==0:
            answer_bata_lavde(o, [])
        elif abs(ord(o[-1])-ord(e[0]))>1:
            answer_bata_lavde(o, e)
        elif abs(ord(o[-1])-ord(e[-1]))>1:
            answer_bata_lavde(o, e[::-1])
        elif abs(ord(o[0])-ord(e[0]))>1:
            answer_bata_lavde(o[::-1], e)
        elif abs(ord(o[0]) - ord(e[-1])) > 1:
            answer_bata_lavde(o[::-1], e[::-1])
        else:
            print('No answer')


    return


if __name__ == "__main__":
    main()