Problem:
Score : 800 points

Problem Statement

You are given integers N,\ A and B.
Determine if there exists a permutation (P_0,\ P_1,\ ...\ P_{2^N-1}) of (0,\ 1,\ ...\ 2^N-1) that satisfies all of the following conditions, and create one such permutation if it exists.

P_0=A

P_{2^N-1}=B

For all 0 \leq i < 2^N-1, the binary representations of P_i and P_{i+1} differ by exactly one bit.

Constraints

1 \leq N \leq 17

0 \leq A \leq 2^N-1

0 \leq B \leq 2^N-1

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

If there is no permutation that satisfies the conditions, print NO.

If there is such a permutation, print YES in the first line.
Then, print (P_0,\ P_1,\ ...\ P_{2^N-1}) in the second line, with spaces in between.
If there are multiple solutions, any of them is accepted.

Sample Input 1

2 1 3

Sample Output 1

YES
1 0 2 3

The binary representation of P=(1,0,2,3) is (01,00,10,11), where any two adjacent elements differ by exactly one bit.

Sample Input 2

3 2 1

Sample Output 2

NO

Solution in Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

program test
implicit none

integer(8) :: n,a,b,count=0,hist = 0,nnn(200000)=0,i,j,gray(200000)=0,ii


read(*,*) n,a,b

do i=0,49
	if(BTEST(a,i) .eqv. BTEST(b,i)) then
		continue
	else
		hist = i
		count = count + 1
	endif
enddo

do i=0,2**n-1
	ii = ishft(i,-1)
	gray(i+1) = ieor(i,ii)
enddo

if(count == 1)then
	write(*,*) "YES"
	
	if(hist == n-1) then
		continue
	else
		do i = 1, 2**n
			if(BTEST(gray(i),n-1) .eqv. BTEST(gray(i),hist) ) then
				continue
			else
				if(BTEST(gray(i),n-1)) then
					gray(i) = ibset(gray(i),hist)
					gray(i) = ibclr(gray(i),n-1)
				else
					gray(i) = ibset(gray(i),n-1)
					gray(i) = ibclr(gray(i),hist)
				endif
			endif
		enddo
	endif
	
	do i =1,2**n
		nnn(i) = ieor(a,gray(i))
	enddo
	
	write(*,*) nnn(1:2**n)/0
	
else
	write(*,*) "NO"
endif

contains
subroutine heapsort_down(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) > y1(j+1)) j = j+1
				endif
				
				if (t1 > y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_down
 
subroutine heapsort_up(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) < y1(j+1)) j = j+1
				endif
				
				if (t1 < y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_up
 
end program