Problem: Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records. 

The records are expressed as a string s of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer p is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer p is considered a period of string s, if for all 1 ≤ i ≤ \lvert s \rvert - p, the i-th and (i + p)-th characters of s are the same. Here \lvert s \rvert is the length of s.

Input

The first line contains two space-separated integers n and p (1 ≤ p ≤ n ≤ 2000) — the length of the given string and the supposed period, respectively.

The second line contains a string s of n characters — Mino's records. s only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that p is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

Input

10 7
1.0.1.0.1.


Output

1000100010


Input

10 6
1.0.1.1000


Output

1001101000


Input

10 9
1........1


Output

No

Note

In the first example, 7 is not a period of the resulting string because the 1-st and 8-th characters of it are different.

In the second example, 6 is not a period of the resulting string because the 4-th and 10-th characters of it are different.

In the third example, 9 is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

Solution (in PYTHON3):

Solution:
n, p = map(int, input().strip().split())
tide = input().strip()

flag = 0

l_tide = list(tide)
rev = {'0':'1', '1':'0'}

for i in range(n-p): # moving through string by index
    # 1  check if char is period.  if so, we can break the period immediately.
    # STRATEGY idea:  just check if there are any periods in checked chars, then can break the pattern.
    if l_tide[i]=='.':
        if l_tide[i+p]=='.': #  this is EASY case, both are wildcards
            l_tide[i] = '0'
            l_tide[i+p] = '1'
            tide_str = ''.join(l_tide)
            tide_str = tide_str.replace('.', '0')
            flag = 1
            print(tide_str)
            break
        else:
            l_tide[i] = rev[l_tide[i+p]] # one is wildcard, switch it to rev of counterpart
            tide_str = ''.join(l_tide)
            tide_str = tide_str.replace('.', '0')
            flag = 1
            print(tide_str)
            break
    else: # the current char is not a period
        # check if remote char is period
        if l_tide[i+p]=='.':  #  break the period here
            l_tide[i+p] = rev[l_tide[i]]
            tide_str = ''.join(l_tide)
            tide_str = tide_str.replace('.', '0')
            flag = 1
            print(tide_str)
            break

        else: # we've just got two numbers.  see if they're the same
            if l_tide[i] == rev[l_tide[i+p]]:
                # period broken
                tide_str = ''.join(l_tide)
                tide_str = tide_str.replace('.', '0')
                flag = 1
                print(tide_str)
                break
            else:
                continue
# if we get to end of loop without breaking, then we can't break period
if flag == 0:
    print('No')