You are making a recipe and need to measure a precise volume
of liquid. There are an assortment of cups of varying volumes
in your kitchen, however no cup has any markings on it other
than to indicate its total volume, and none of them match the
volume that you want. You start with the biggest cup full of
liquid and, to make sure you know precisely how much volume you
are working with at any point in time, you consider steps in
which you pour from any nonempty cup into another cup, always
pouring until either the cup you are pouring into becomes full,
or the cup you are pouring from becomes empty (whichever occurs
first). As a simple example, assume you start with a full cup
having capacity $5$,
and you have another cup with capacity $2$, but your goal is to have
$3$ units of the
liquid in the largest cup. In this case, you can start pouring
from the larger cup to the smaller, stopping when the smaller
one reaches its capacity of $2$. This will leave precisely
$3$ units in the
larger cup. See Figure 1(a).
As another example, consider a case in which you have
$4$ cups with capacities
$9$, $6$, $3$, and $2$, and you start with the largest
cup full, the rest empty, and a goal of ending with
$8$ units in the
largest cup. For ease of discussion we will refer to the cup
with capacity $9$ as
the “$9$cup” and
similarly for the other sizes. You notice that the $6$cup and $2$cup have combined capacity of
$8$, and so you could pour
from the original $9$cup
to fill those two cups, then dump the remaining $1$ unit from the $9$cup into the $3$cup, and finally pour the full
$6$cup and $2$cup back into the $9$cup. See Figure 1(b). In
implementing this strategy, the total volume of liquid poured
would be $6+2+1+6+2=17$.
You could achieve this goal in another way: pour $3$ units from the $9$cup to the $3$cup (leaving $6$ units in the $9$cup), then fill the $2$cup from the $3$cup (leaving $1$ unit in the $2$ cup), and finally pour the
full $2$cup back into the
$9$cup, resulting in
exactly $8$ units in
that cup. With this strategy, the total volume poured is only
$3+2+2=7$. See
Figure 1(c).
As a final example, you start with cups of capacities
$11$, $10$, $7$, $4$, and $2$, with the $11$cup full, and a goal of ending up
with $10$ units in
the $11$cup. Obviously,
you could fill the $10$cup, dump the remaining
$1$ unit into another
cup, and then pour from the full $10$cup back into the $11$cup, as illustrated in
Figure 2(a). These three pours would mean transferring a
total volume of $10+1+10=21$. Figure 2(b) shows a
sequence with more steps, but less liquid poured.
Input
The input consists of a single line of positive integers:
$n$ $c_1$ $c_2$ $\ldots $ $c_ n$ $V$, where there are $n$ cups, with $2 \leq n \leq 5$, having capacities
satisfying $ 64 \geq
c_1>c_2> \ldots > c_ n \geq 1$. The value
$V < c_1$ designates
the desired volume. You must start with largest cup (that with
capacity $c_1$) full
of liquid and the other cups empty, and the goal is to get
exactly volume $V$
into the largest cup.
Output
Output the minimum amount of liquid that must be poured to
achieve the goal, or output impossible if
the goal cannot be achieved.
Sample Input 1 
Sample Output 1 
2 5 2 3

2

Sample Input 2 
Sample Output 2 
4 9 6 3 2 8

7

Sample Input 3 
Sample Output 3 
5 11 10 7 4 2 10

19

Sample Input 4 
Sample Output 4 
2 5 2 4

impossible

Sample Input 5 
Sample Output 5 
5 64 45 41 28 2 63

121
