Problem: Acacius is studying strings theory. Today he came with the following problem.

You are given a string s of length n consisting of lowercase English letters and question marks. It is possible to replace question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once?

Each question mark should be replaced with exactly one lowercase English letter. For example, string "a?b?c" can be transformed into strings "aabbc" and "azbzc", but can't be transformed into strings "aabc", "a?bbc" and "babbc".

Occurrence of a string t of length m in the string s of length n as a substring is a index i (1 ≤ i ≤ n - m + 1) such that string s[i..i+m-1] consisting of m consecutive symbols of s starting from i-th equals to string t. For example string "ababa" has two occurrences of a string "aba" as a substring with i = 1 and i = 3, but there are no occurrences of a string "aba" in the string "acba" as a substring.

Please help Acacius to check if it is possible to replace all question marks with lowercase English letters in such a way that a string "abacaba" occurs as a substring in a resulting string exactly once.

Input

First line of input contains an integer T (1 ≤ T ≤ 5000), number of test cases. T pairs of lines with test case descriptions follow.

The first line of a test case description contains a single integer n (7 ≤ n ≤ 50), length of a string s.

The second line of a test case description contains string s of length n consisting of lowercase English letters and question marks.

Output

For each test case output an answer for it.

In case if there is no way to replace question marks in string s with a lowercase English letters in such a way that there is exactly one occurrence of a string "abacaba" in the resulting string as a substring output "No".

Otherwise output "Yes" and in the next line output a resulting string consisting of n lowercase English letters. If there are multiple possible strings, output any.

You may print every letter in "Yes" and "No" in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answer).

Example

Input


6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba


Output


Yes
abacaba
Yes
abacaba
Yes
abadabacaba
Yes
abacabadaba
No
No

Note

In first example there is exactly one occurrence of a string "abacaba" in the string "abacaba" as a substring.

In second example seven question marks can be replaced with any seven lowercase English letters and with "abacaba" in particular.

In sixth example there are two occurrences of a string "abacaba" as a substring.

Solution (in PYTHON3):

Solution:
"""Template for Python Competitive Programmers prepared by pajengod and   many others """
#  ////////// SHUBHAM SHARMA \\\\\\\\\\\\\
# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10 ** 9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""


def prepare_factorial():
    fact = [1]
    for i in range(1, 100005):
        fact.append((fact[-1] * i) % mod)
    ifact = [0] * 100005
    ifact[100004] = pow(fact[100004], mod - 2, mod)
    for i in range(100004, 0, -1):
        ifact[i - 1] = (i * ifact[i]) % mod
    return fact, ifact


"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(1<<27)
import sys

# sys.setrecursionlimit(10000)


"""uncomment modules according to your need"""
from bisect import bisect_left, bisect_right, insort
# from itertools import repeat
from math import floor, ceil, sqrt, degrees, atan, pi, log, sin, radians
from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
from collections import Counter, defaultdict, deque


# from copy import deepcopy
# from decimal import *
# import re
# import operator


def modinv(n, p):
    return pow(n, p - 2, p)


def ncr(n, r, fact, ifact):  # for using this uncomment the lines calculating fact and ifact
    t = (fact[n] * (ifact[r] * ifact[n - r]) % mod) % mod
    return t


def intarray(): return map(int, sys.stdin.readline().strip().split())


def array(): return list(map(int, sys.stdin.readline().strip().split()))


def input(): return sys.stdin.readline().strip()


"""*****************************************************************************************"""


def GCD(x, y):
    while y:
        x, y = y, x % y
    return x


def lcm(x, y):
    return (x * y) // (GCD(x, y))


def get_xor(n):
    return [n, 1, n + 1, 0][n % 4]


def fast_expo(a, b):
    res = 1
    while b:
        if b & 1:
            res = (res * a)
            res %= MOD
            b -= 1
        else:
            a = (a * a)
            a %= MOD
            b >>= 1
    res %= MOD
    return res


def get_n(P):  # this function returns the maximum n for which Summation(n) <= Sum

    ans = (-1 + sqrt(1 + 8 * P)) // 2
    return ans


""" ********************************************************************************************* """
"""
 array() # for araay
 intarray() # for map array
 SAMPLE INPUT HERE
6
7
abacaba
7
???????
11
aba?abacaba
11
abacaba?aba
15
asdf???f???qwer
11
abacabacaba


"""
""" OM SAI RAM """
cur = []
req = 'abacaba'
def check(ind, s,n):
    global cur
    cur = list(s)
    for i in range(0, 7):
        if cur[ind+i] == '?':
            cur[ind+i] = req[i]
        elif cur[ind+i] != s[ind+i]:
            return 0
    for i in range(0,len(s)):
        if cur[i] == '?':
            cur[i] = 'z'
    count = 0
    for i in range(n - 6):
        for j in range(0, 7):
            if req[j] != cur[i+j]:
                break
        else:
            count += 1
    if count == 1:
        return True
    else:
        return False
def solve():
    n = int(input())
    s = input()
    for i in range(n - 6):
        if check(i,s,n):
            print('YES')
            print(''.join(cur))
            return
    print('NO')
    return

def main():
    T = int(input())
    while T:
        solve()
        T -= 1


"""OM SAI RAM """


""" -------- Python 2 and 3 footer by Pajenegod and c1729 ---------"""

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

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


class IOWrapper(IOBase):

    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

""" main function"""

if __name__ == '__main__':
    main()
    # threading.Thread(target=main).start()