Problem: E: Red Black Balloons

Story

Homura-chan's dream comes true. It means ICPC Asia regional contest 20xx will be held in Sapporo! Homura-chan has been working hard for the preparation. And finally, it's the previous day of the contest. Homura-chan started to stock balloons to be delivered to contestants who get accepted. However, she noticed that there were only two colors of balloons: red and black.

Problem Statement

ICPC Asia regional contest in Sapporo plans to provide N problems to contestants. Homura-chan is a professional of contest preparation, so she already knows how many contestants would get acceptance for each problem (!!), a_i contestants for the i-th problem. You can assume the prediction is perfectly accurate. Ideally, Homura-chan would assign a distinct color of a balloon to each problem respectively. But you know, she has only two colors red and black now. Thus, Homura-chan comes up with the idea to differentiate the size of balloons in K levels, that is, each problem has a balloon with a distinct pair of (color, size).

Homura-chan now has r_i red balloons with size i (1 \leq i \leq K) and b_j black balloons with size j (1 \leq j \leq K). Suppose we assign a pair (c_i, s_i) of a color c_i (red or black) and a size s_i to the i-th problem, for each i. As noted, we have to assign distinct pairs to each problem, more precisely, (c_i, s_i) \neq (c_j, s_j) holds for i \neq j. Moreover, the number of balloons with (c_i, s_i) must be no less than a_i. In the case that there are no such assignments, Homura-chan can change the size of balloons by her magic power. Note that Homura-chan doesn't know magic to change the color of balloons, so it's impossible.

Your task is to write a program computing the minimum number of balloons whose size is changed by Homura-chan's magic to realize an assignment satisfying the above-mentioned conditions. If there is no way to achieve such an assignment even if Homura-chan changes the size of balloons infinitely, output `-1` instead.

Input


N K
a_1 ... a_N
r_1 ... r_K
b_1 ... b_K


Constraints

* 1 \leq N,K \leq 60
* N \leq 2K
* 1 \leq a_i \leq 50 (1 \leq i \leq N)
* 1 \leq r_j,b_j \leq 50 (1 \leq j \leq K)
* Inputs consist only of integers.



Output

Output the minimum number of balloons whose size is changed to achieve an assignment in a line. If there are no ways to achieve assignments, output `-1` instead.

Sample Input 1


3 2
6 5 4
8 1
7 1


Output for Sample Input 1


3

Homura-chan changes the size of three red balloons from 1 to 2. Then she can assign (black,1) to the problem 1, (red,1) to the problem 2, and (red,2) to the problem 3.

Sample Input 2


2 1
50 50
2
3


Output for Sample Input 2


-1

Sample Input 3


4 3
3 10 28 43
40 18 2
26 7 11


Output for Sample Input 3


5





Example

Input

3 2
6 5 4
8 1
7 1


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n,m,a[109],r[109],b[109],sum[109],sumr,sumb;
bool cmp(int x,int y){return x>y;}
int f[69][69][3009];

//f[i][j][k]表示r匹配到i，b匹配到j，与r匹配的a之和为k 

int main() {
	scanf("%d%d",&n,&m);sum[0]=sumr=sumb=0;
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=m;i++) scanf("%d",&r[i]);
	for(int i=1;i<=m;i++) scanf("%d",&b[i]);
	sort(a+1,a+n+1,cmp);
	sort(r+1,r+m+1,cmp);sort(b+1,b+m+1,cmp);
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
	for(int i=1;i<=m;i++) sumr+=r[i],sumb+=b[i];
	memset(f,-1,sizeof(f));
	f[0][0][0]=0;int ans=-1;
	for(int i=0;i<=m;i++){
		for(int j=0;j<=m;j++){
			if(i+j>n) continue;
			for(int k=0;k<=sum[i+j];k++){
				if(f[i][j][k]==-1) continue;
				if(i+j==n){
					if(sumr>=k&&sumb>=sum[n]-k) ans=max(ans,f[i][j][k]+sumr+sumb-sum[n]);
					continue;
				} 
				if(sumr>=k+a[i+j+1]&&i<m) f[i+1][j][k+a[i+j+1]]=max(f[i+1][j][k+a[i+j+1]],f[i][j][k]+min(a[i+j+1],r[i+1]));
				if(sumb>=sum[i+j+1]-k&&j<m) f[i][j+1][k]=max(f[i][j+1][k],f[i][j][k]+min(a[i+j+1],b[j+1])); 
			}
		}
	}
	printf("%d\n",ans==-1?ans:(sumr+sumb-ans));	
	return 0;
}