Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.

Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.

Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
The first line contains a single positive integer *n* (1 ≤ *n* ≤ 100) — the length of the canvas.

The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.

For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.

For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.

For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
5
CY??Y
Yes
5
C?C?Y
Yes
5
?CYC?
Yes
5
C??MM
No
3
MMY
No
15
??YYYYYY??YYYY?
No
100
YCY?CMCMCYMYMYC?YMYMYMY?CMC?MCMYCMYMYCM?CMCM?CMYMYCYCMCMCMCMCMYM?CYCYCMCM?CY?MYCYCMYM?CYCYCYMY?CYCYC
No
1
C
No
1
?
Yes
2
MY
No
2
?M
Yes
2
Y?
Yes
2
??
Yes
3
??C
Yes
3
M??
Yes
3
YCM
No
3
?C?
Yes
3
MC?
Yes
4
CYCM
No
4
M?CM
No
4
??YM
Yes
4
C???
Yes
10
MCYM?MYM?C
Yes
50
CMCMCYM?MY?C?MC??YM?CY?YM??M?MCMCYCYMCYCMCM?MCM?MC
Yes
97
MCM?YCMYM?YMY?MY?MYCY?CMCMCYC?YMY?MYCMC?M?YCMC?YM?C?MCMCMYMCMY?MCM?YC?YMYMY?MYCYCM?YC?YCY?MYMYMYC
No
100
C?M?M?M?YM??YMYC?MCYMYM??Y??YC?CYC???YM?YM??MYMY?CYCYMYC?YC?C?CYCMY??CMC?YMCMYCYCYMYM?CYM?M?MCMCMY?Y
Yes
100
?YYYYYYYYYYYYYYYYYYYYYYYYYYYYY??YYY?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY?
No
100
????????????????????????????????????????????????????????????????????????????????????????????????????
Yes
100
Y?CYMYMYMYCYMY?CMCYMYMC?MYCY?CMCYCYMCM?YCMCYMYCYCY?MYMYCYCYCM?CYMYMY?CY?MYCYCY?M?YMYCY?CMCMCY?CY?M?C
No
4
M??C
Yes
6
Y?C??C
Yes
5
C???Y
Yes
5
C??MY
Yes
5
CY??M
Yes
4
C??Y
Yes
52
??????????????????????????????????????????????????YY
No
3
YY?
No
5
CC??Y
No
8
CMC??MCM
Yes
7
M?YCM??
Yes
6
?CC???
No
100
??????????????????????????????????????????????????????????????????????????????????????????????????MM
No
4
C??M
Yes
4
?C?M
Yes
6
MC??MC
Yes
"""
Usage:
    python solution.py < input.txt > solution_output.txt

Testing:
    pytest test_solution.py

Submit:
    Just copy and paste all the code or upload the whole file
"""
from itertools import product


def get_input_list() -> str:
    """
    Loops throw the stdio to fetch input data
    Generally used on problems with 'n' inputs

    Returns:
        raw_input_data: The input of a problem
    """
    canvas_lenght = input()
    canvas_caracters = input()
        
    return canvas_caracters
    

def get_input() -> str:
    """
    Fetch input data from stdio
    Generally used on problems with discrete inputs
    Returns:
        raw_input_data: The input of a problem
    """
    raw_input_data = ""
    raw_input_data += input()
    return raw_input_data


def solve(raw_input_data: str):
    """
    Here you solve the problem, happy coding!!
    Args:
        raw_input_data: the data you test the solution

    Returns:
        answer: the solution of the problem
    """
    word = raw_input_data.split()[0]

    combinations = list(product("CMY", repeat=5))

    # From all combinations returned by de product,
    # saves only the ones that do not have repeated
    # elements
    aux_list = []
    for com in combinations:
        for i in range(len(com)):
            if com[i] == com[i - 1]:
                break
        else:
            aux_list.append(com)

    for comb in aux_list:
        # Loops through all letters in word
        # and check if they're equal to comb's
        # If one is correct, print Yes, otherwise
        # print no
        for c1, c2 in zip(word, comb):
            if c1 != "?" and c1 != c2:
                break
        else:
            print("Yes", end="\n")
            break
    else:
        print("No", end="\n")


if __name__ == "__main__":
    
    while True:
        try:
            # Use this with simple inputs
            # input_data = get_input()
            
            # Use this with the problem have many inputs
            input_data = get_input_list()
            
            answer = solve(input_data)
            # print(answer, end="\n")
        except:
            break
