program test
implicit none

integer(8),parameter :: inf_minus = -1d10
integer(8) :: i,j,k,l,m,n
integer(8) :: a(10),b(10),dp(-100:10000)=inf_minus
integer(8) :: match_use(9) = (/2,5,5,4,5,6,3,7,6/)
integer(8) :: ans(10000)=0,rest

read(*,*) n,m
read(*,*) (a(i),i=1,m)
do i=1,m
	b(i) = match_use(a(i))
enddo

call heapsort_down(m,a(1:m),b(1:m))

dp(0) = 0

!dp(i)とは、i本のマッチを使って作成できる最大の文字数
do i=0,n
	do j=1,m
		if(i-b(j)>=0 )then
			dp(i) = max(dp(i),(dp(i-b(j))+1))
		endif
	enddo
enddo

!上の桁から、利用可能な数字か確かめていく
rest = n
do i=1,dp(n)
	do j=1,m
		if(rest-b(j)>=0 .and.dp(rest-b(j))==(dp(rest)-1))then
			rest = rest-b(j)
			ans(i) = a(j)
			goto 100
		endif
	enddo
	100 continue
enddo


write(*,'(10000(I1))') (ans(i),i=1,dp(n))

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