p03111 AtCoder Beginner Contest 119 - Synthetic Kadomatsu
You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

* Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.
* Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.
* Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)



At least how much MP is needed to achieve the objective?

Constraints

* 3 \leq N \leq 8
* 1 \leq C < B < A \leq 1000
* 1 \leq l_i \leq 1000
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N A B C
l_1
l_2
:
l_N


Output

Print the minimum amount of MP needed to achieve the objective.

Examples

Input

5 100 90 80
98
40
30
21
80


Output

23


Input

8 100 90 80
100
100
90
90
90
80
80
80


Output

0


Input

8 1000 800 100
300
333
400
444
500
555
600
666


Output

243
```python
n,a,b,c=map(int,input().split())
l=[int(input()) for _ in range(n)]

def dfs(i,x,y,z):
  if i==n:
    return abs(x-a)+abs(y-b)+abs(z-c) if x*y*z else 10**9
  r1 = dfs(i+1,x+l[i],y,z)+ (10 if x>0 else 0)
  r2 = dfs(i+1,x,y+l[i],z)+ (10 if y>0 else 0)
  r3 = dfs(i+1,x,y,z+l[i])+ (10 if z>0 else 0)
  r4 = dfs(i+1,x,y,z)
  return min(r1,r2,r3,r4)

print(dfs(0,0,0,0))

```