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>

#define pb push_back
#define mp make_pair
#define fi first
#define se second

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

template <typename T> bool chkmax(T &x,T y){return x<y?x=y,true:false;}
template <typename T> bool chkmin(T &x,T y){return x>y?x=y,true:false;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,k,sr,sb,sa;
int a[65],r[65],b[65],d[65][3005],f[65][3005];

int main(){
	n=readint(); k=readint();
	for(int i=1;i<=n;i++) a[i]=readint();
	for(int i=1;i<=k;i++) r[i]=readint();
	for(int i=1;i<=k;i++) b[i]=readint();
	for(int i=1;i<=k;i++) sr+=r[i],sb+=b[i];
	for(int i=1;i<=n;i++) sa+=a[i];
	sort(r+1,r+k+1); reverse(r+1,r+k+1);
	sort(b+1,b+k+1); reverse(b+1,b+k+1);
	sort(a+1,a+n+1); reverse(a+1,a+n+1);
	memset(f,0x3f,sizeof(f));
	memset(d,0x3f,sizeof(d));
	d[0][0]=0;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=i;j++){
			for(int l=0;l<=sr;l++){
				if(j&&j<=k&&l>=a[i]) chkmin(f[j][l],d[j-1][l-a[i]]+max(0,a[i]-r[j]));
				if(i-j&&i-j<=k) chkmin(f[j][l],d[j][l]+max(0,a[i]-b[i-j]));
			}
		}
		swap(d,f);
		memset(f,0x3f,sizeof(f));
	}
	int ans=0x3f3f3f3f;
	for(int i=0;i<=k;i++) for(int k=0;k<=sr;k++) if(sa-k<=sb) chkmin(ans,d[i][k]);
	printf("%d\n",ans==(0x3f3f3f3f)?-1:ans);
	return 0;
}